#include <bits/stdc++.h>
#define pi acos(-1.0)
// #define double long double
using namespace std;
typedef long long LL;
const double eps = 1e-8, inf = 1e12;
const int N = 1e2 + 10;
int sgn(double x)
{
    if (fabs(x) < eps) return 0;
    if (x < 0) return -1;
    else return 1;
}
int dsgn(double x, double y)
{
    if (fabs(x - y) < eps) return 0;
    if (x < y) return -1;
    else return 1;
}
inline double sqr(double x) { return x * x; }

struct Point
{
    double x, y;
    Point() {}
    Point(double _x, double _y) { x = _x;y = _y; }
    void input() { cin >> x >> y; }
    bool operator==(Point b) const { return sgn(x - b.x) == 0 && sgn(y - b.y) == 0; }
    bool operator<(Point b) const { return sgn(x - b.x) == 0 ? sgn(y - b.y) < 0 : x < b.x; }
    Point operator-(const Point &b) const { return Point(x - b.x, y - b.y); }
    Point operator+(const Point &b) const { return Point(x + b.x, y + b.y); }
    // 叉积
    double operator^(const Point &b) const { return x * b.y - y * b.x; }
    // 点积
    double operator*(const Point &b) const { return x * b.x + y * b.y; }
    // 数乘
    Point operator*(const double &k) const {return Point(x * k, y * k);}
    Point operator/(const double &k) const { return Point(x / k, y / k);}
    // 返回长度
    double len() { return hypot(x, y); }
    // 返回长度的平方
    double len2() { return x * x + y * y; }
    // 返回两点的距离
    double distance(Point p) { return hypot(x - p.x, y - p.y); }
    LL distance2(Point p) { return (x - p.x) * (x - p.x) + (y - p.y) * (y - p.y); }
    //`计算pa  和  pb 的夹角 就是求这个点看a,b 所成的夹角`
    //`测试 LightOJ1203`
    double rad(Point a, Point b)
    {
        Point p = *this;
        return fabs(atan2(fabs((a - p) ^ (b - p)), (a - p) * (b - p)));
    }
    //`绕着p点逆时针旋转angle`
    Point rotate(Point p, double angle)
    {
        Point v = (*this) - p;
        double c = cos(angle), s = sin(angle);
        return Point(p.x + v.x * c - v.y * s, p.y + v.x * s + v.y * c);
    }
    //`化为长度为r的向量`
    Point trunc(double r)
    {
        double l = len();
        if (!sgn(l)) return *this;
        r /= l;
        return Point(x * r, y * r);
    }
    //`逆时针旋转90度`
    Point rotleft() { return Point(-y, x); }
    //`顺时针旋转90度`;
    Point rotright() { return Point(y, -x); }
};
//`AB X AC`
double cross(Point A, Point B, Point C) { return (B - A) ^ (C - A); }
//`AB*AC`
double dot(Point A, Point B, Point C) { return (B - A) * (C - A); }
struct Line
{
    Point s, e;
    Line() {}
    Line(Point _s, Point _e) { s = _s;e = _e; }
    bool operator==(Line v) { return (s == v.s) && (e == v.e); }
    //`根据一个点和倾斜角angle确定直线,0<=angle<pi`
    Line(Point p, double angle)
    {
        s = p;
        if (sgn(angle - pi / 2) == 0) e = (s + Point(0, 1));
        else e = (s + Point(1, tan(angle)));
    }
    // ax+by+c=0
    Line(double a, double b, double c)
    {
        if (sgn(a) == 0)
        {
            s = Point(0, -c / b);
            e = Point(1, -c / b);
        }
        else if (sgn(b) == 0)
        {
            s = Point(-c / a, 0);
            e = Point(-c / a, 1);
        }
        else
        {
            s = Point(0, -c / b);
            e = Point(1, (-c - a) / b);
        }
    }
    void input() { s.input();e.input(); }
    void adjust() { if (e < s) swap(s, e); }
    // 求线段长度
    double length() { return s.distance(e); }
    //`返回直线倾斜角 0<=angle<pi`
    double angle()
    {
        double k = atan2(e.y - s.y, e.x - s.x);
        // if(sgn(k) < 0)k += pi;
        // if(sgn(k-pi) == 0)k -= pi;
        return k;
    }
    bool operator<(Line &v) { return sgn(angle() - v.angle()) == 0 ? ((e - s) ^ (v.e - s)) < 0 : angle() < v.angle(); }
    //`点和直线关系`
    int relation(Point p)
    {
        int c = sgn((p - s) ^ (e - s));
        if (c < 0) return 1;      //`1  在左侧`
        else if (c > 0) return 2; //`2  在右侧`
        else return 3;            //`3  在直线上`
    }
    //`点在线段上的判断`
    bool pointonseg(Point p) { return sgn((p - s) ^ (e - s)) == 0 && sgn((p - s) * (p - e)) <= 0; }
    //`两向量平行(对应直线平行或重合)`
    bool parallel(Line v) { return sgn((e - s) ^ (v.e - v.s)) == 0; }
    //`两向量正交`
    bool orthogonal(Line v) { return sgn((e - s) * (v.e - v.s)) == 0; }
    //`两直线关系`
    int linecrossline(Line v)
    {
        if ((*this).parallel(v))             //`0 平行`
            return v.relation(s) == 3;       //`1 重合`
        if ((*this).orthogonal(v)) return 3; //`3 正交`
        else return 2;                       //`2 相交`
    }
    //`两线段相交判断`
    int segcrossseg(Line v)
    {
        int d1 = sgn((e - s) ^ (v.s - s));
        int d2 = sgn((e - s) ^ (v.e - s));
        int d3 = sgn((v.e - v.s) ^ (s - v.s));
        int d4 = sgn((v.e - v.s) ^ (e - v.s));
        if ((d1 ^ d2) == -2 && (d3 ^ d4) == -2) return 2;      //`2 规范相交`
        return (d1 == 0 && sgn((v.s - s) * (v.s - e)) <= 0) || //`1 非规范相交`
               (d2 == 0 && sgn((v.e - s) * (v.e - e)) <= 0) || //`0 不相交`
               (d3 == 0 && sgn((s - v.s) * (s - v.e)) <= 0) ||
               (d4 == 0 && sgn((e - v.s) * (e - v.e)) <= 0);
    }
    //`直线和线段相交判断`
    int linecrossseg(Line v)
    {
        int d1 = sgn((e - s) ^ (v.s - s));
        int d2 = sgn((e - s) ^ (v.e - s));
        if ((d1 ^ d2) == -2) return 2;     //`2 规范相交`
        return (d1 == 0 || d2 == 0);       //`1 非规范相交 0 不相交`
    }
    //`求两直线的交点  要保证两直线不平行或重合`
    Point crosspoint(Line l)
    {
        Point u = e - s, v = l.e - l.s;
        double t = (s - l.s) ^ v / (v ^ u);
        return s + u * t;
    }
    //`返回点p在直线上的投影`
    Point lineprog(Point p)
    {
        Point u = e - s, v = p - s;
        return s + u * (u * v / u.len2());
    }
    // 点到直线的距离
    double dispointtoline(Point p) { return fabs((p - s) ^ (e - s)) / length(); }
    // 点到线段的距离
    double dispointtoseg(Point p)
    {
        if (sgn((p - s) * (e - s)) < 0 || sgn((p - e) * (s - e)) < 0)
            return min(p.distance(s), p.distance(e));
        return dispointtoline(p);
    }
    //`返回线段到线段的距离 前提是两线段不相交，相交距离就是0了`
    double dissegtoseg(Line v)
    {
        return min(min(dispointtoseg(v.s), dispointtoseg(v.e)), min(v.dispointtoseg(s), v.dispointtoseg(e)));
    }
    //`返回点p关于直线的对称点`
    Point symmetrypoint(Point p)
    {
        Point pro = lineprog(p);
        return pro * 2 - p;
    }
};

struct circle
{
    Point p;  // 圆心
    double r; // 半径
    circle() {}
    circle(Point _p, double _r) { p = _p;r = _r; }
    circle(double x, double y, double _r) { p = Point(x, y);r = _r; }
    //`三角形的外接圆`
    //`需要Point的+ /  rotate()  以及Line的crosspoint()`
    //`利用两条边的中垂线得到圆心`
    //`测试：UVA12304`
    circle(Point a, Point b, Point c)
    {
        Line u = Line((a + b) / 2, ((a + b) / 2) + ((b - a).rotleft()));
        Line v = Line((b + c) / 2, ((b + c) / 2) + ((c - b).rotleft()));
        p = u.crosspoint(v);
        r = p.distance(a);
    }
    //`三角形的内切圆`
    //`参数bool t没有作用，只是为了和上面外接圆函数区别`
    //`测试：UVA12304`
    circle(Point a, Point b, Point c, bool t)
    {
        Line u, v;
        double m = atan2(b.y - a.y, b.x - a.x), n = atan2(c.y - a.y, c.x - a.x);
        u.s = a;
        u.e = u.s + Point(cos((n + m) / 2), sin((n + m) / 2));
        v.s = b;
        m = atan2(a.y - b.y, a.x - b.x), n = atan2(c.y - b.y, c.x - b.x);
        v.e = v.s + Point(cos((n + m) / 2), sin((n + m) / 2));
        p = u.crosspoint(v);
        r = Line(a, b).dispointtoseg(p);
    }
    // 输入
    void input() { p.input();cin >> r; }
    bool operator==(circle v) { return (p == v.p) && sgn(r - v.r) == 0; }
    bool operator<(circle v) const { return ((p < v.p) || ((p == v.p) && sgn(r - v.r) < 0)); }
    // 面积
    double area() { return pi * r * r; }
    // 周长
    double circumference() { return 2 * pi * r; }
    //`点和圆的关系`
    int relation(Point b)
    {
        double dst = b.distance(p);
        if (sgn(dst - r) < 0) return 2;       //`2 圆内`
        else if (sgn(dst - r) == 0) return 1; //`1 圆上`
        return 0;                             //`0 圆外`
    }
    //`线段和圆的关系 比较的是圆心到线段的距离和半径的关系`
    int relationseg(Line v)
    {
        double dst = v.dispointtoseg(p);
        if (sgn(dst - r) < 0) return 2;       //`2 相交`
        else if (sgn(dst - r) == 0) return 1; //`1 相切`
        return 0;                             //`0 相离`
    }
    //`直线和圆的关系 比较的是圆心到直线的距离和半径的关系`
    int relationline(Line v)
    {
        double dst = v.dispointtoline(p);
        if (sgn(dst - r) < 0) return 2;       //`2 相交`
        else if (sgn(dst - r) == 0) return 1; //`1 相切`
        return 0;                             //`0 相离`
    }
    //`两圆的关系 需要Point的distance`
    //`测试：UVA12304`
    int relationcircle(circle v)
    {
        double d = p.distance(v.p);
        if (sgn(d - r - v.r) > 0) return 5;                   //`5 相离`
        if (sgn(d - r - v.r) == 0) return 4;                  //`4 外切`
        double l = fabs(r - v.r);
        if (sgn(d - r - v.r) < 0 && sgn(d - l) > 0) return 3; //`3 相交`
        if (sgn(d - l) == 0) return 2;                        //`2 内切`
        if (sgn(d - l) < 0) return 1;                         //`1 内含`
    }
    //`求两个圆的交点 需要relationcircle`
    //`测试：UVA12304`
    int pointcrosscircle(circle v, Point &p1, Point &p2)
    {
        int rel = relationcircle(v);
        if (rel == 1 || rel == 5) return 0;               //`0 无交点`
        double d = p.distance(v.p);
        double l = (d * d + r * r - v.r * v.r) / (2 * d);
        double h = sqrt(r * r - l * l);
        Point tmp = p + (v.p - p).trunc(l);
        p1 = tmp + ((v.p - p).rotleft().trunc(h));
        p2 = tmp + ((v.p - p).rotright().trunc(h));
        if (rel == 2 || rel == 4) return 1;               //`1 一个交点`
        return 2;                                         //`2 两个交点`
    }
    //`求直线和圆的交点，返回交点个数`
    int pointcrossline(Line v, Point &p1, Point &p2)
    {
        if (!(*this).relationline(v)) return 0;
        Point a = v.lineprog(p);
        double d = v.dispointtoline(p);
        d = sqrt(r * r - d * d);
        if (sgn(d) == 0)
        {
            p1 = a;
            p2 = a;
            return 1;
        }
        p1 = a + (v.e - v.s).trunc(d);
        p2 = a - (v.e - v.s).trunc(d);
        return 2;
    }
    //`得到过a,b两点，半径为r1的两个圆`
    int gercircle(Point a, Point b, double r1, circle &c1, circle &c2)
    {
        circle x(a, r1), y(b, r1);
        int t = x.pointcrosscircle(y, c1.p, c2.p);
        if (!t) return 0;
        c1.r = c2.r = r;
        return t; //`返回圆的个数`
    }
    //`得到与直线u相切，过点q,半径为r1的圆`
    //`测试：UVA12304`
    int getcircle(Line u, Point q, double r1, circle &c1, circle &c2)
    {
        double dis = u.dispointtoline(q);
        if (sgn(dis - r1 * 2) > 0) return 0;
        if (sgn(dis) == 0)
        {
            c1.p = q + ((u.e - u.s).rotleft().trunc(r1));
            c2.p = q + ((u.e - u.s).rotright().trunc(r1));
            c1.r = c2.r = r1;
            return 2;
        }
        Line u1 = Line((u.s + (u.e - u.s).rotleft().trunc(r1)), (u.e + (u.e - u.s).rotleft().trunc(r1)));
        Line u2 = Line((u.s + (u.e - u.s).rotright().trunc(r1)), (u.e + (u.e - u.s).rotright().trunc(r1)));
        circle cc = circle(q, r1);
        Point p1, p2;
        if (!cc.pointcrossline(u1, p1, p2)) cc.pointcrossline(u2, p1, p2);
        c1 = circle(p1, r1);
        if (p1 == p2)
        {
            c2 = c1;
            return 1;
        }
        c2 = circle(p2, r1);
        return 2;
    }
    //`同时与直线u,v相切，半径为r1的圆`
    //`测试：UVA12304`
    int getcircle(Line u, Line v, double r1, circle &c1, circle &c2, circle &c3, circle &c4)
    {
        if (u.parallel(v)) return 0; // 两直线平行
        Line u1 = Line(u.s + (u.e - u.s).rotleft().trunc(r1), u.e + (u.e - u.s).rotleft().trunc(r1));
        Line u2 = Line(u.s + (u.e - u.s).rotright().trunc(r1), u.e + (u.e - u.s).rotright().trunc(r1));
        Line v1 = Line(v.s + (v.e - v.s).rotleft().trunc(r1), v.e + (v.e - v.s).rotleft().trunc(r1));
        Line v2 = Line(v.s + (v.e - v.s).rotright().trunc(r1), v.e + (v.e - v.s).rotright().trunc(r1));
        c1.r = c2.r = c3.r = c4.r = r1;
        c1.p = u1.crosspoint(v1);
        c2.p = u1.crosspoint(v2);
        c3.p = u2.crosspoint(v1);
        c4.p = u2.crosspoint(v2);
        return 4;
    }
    //`同时与不相交圆cx,cy相切，半径为r1的圆`
    //`测试：UVA12304`
    int getcircle(circle cx, circle cy, double r1, circle &c1, circle &c2)
    {
        circle x(cx.p, r1 + cx.r), y(cy.p, r1 + cy.r);
        int t = x.pointcrosscircle(y, c1.p, c2.p);
        if (!t) return 0;
        c1.r = c2.r = r1;
        return t; //`返回圆的个数`
    }
    //`过一点作圆的切线(先判断点和圆的关系)`
    //`测试：UVA12304`
    int tangentline(Point q, Line &u, Line &v)
    {
        int x = relation(q);
        if (x == 2)  return 0;
        if (x == 1)
        {
            u = Line(q, q + (q - p).rotleft());
            v = u;
            return 1;
        }
        double d = p.distance(q);
        double l = r * r / d;
        double h = sqrt(r * r - l * l);
        u = Line(q, p + ((q - p).trunc(l) + (q - p).rotleft().trunc(h)));
        v = Line(q, p + ((q - p).trunc(l) + (q - p).rotright().trunc(h)));
        return 2; //`返回切线的个数`
    }
    int tangentpoint(Point q, Point &u, Point &v)
    {
        int x = relation(q);
        if (x == 2) return 0;
        if (x == 1)
        {
            u = q;
            v = u;
            return 1;
        }
        double d = p.distance(q);
        double l = r * r / d;
        double h = sqrt(r * r - l * l);
        u = p + ((q - p).trunc(l) + (q - p).rotleft().trunc(h));
        v = p + ((q - p).trunc(l) + (q - p).rotright().trunc(h));
        return 2; //`返回切点的个数`
    }
    //`求两圆相交的面积`
    double areacircle(circle v)
    {
        int rel = relationcircle(v);
        if (rel >= 4) return 0.0;
        if (rel <= 2) return min(area(), v.area());
        double d = p.distance(v.p);
        double hf = (r + v.r + d) / 2.0;
        double ss = 2 * sqrt(hf * (hf - r) * (hf - v.r) * (hf - d));
        double a1 = acos((r * r + d * d - v.r * v.r) / (2.0 * r * d));
        a1 = a1 * r * r;
        double a2 = acos((v.r * v.r + d * d - r * r) / (2.0 * v.r * d));
        a2 = a2 * v.r * v.r;
        return a1 + a2 - ss;
    }
    //`求圆和三角形pab的相交面积`
    //`测试：POJ3675 HDU3982 HDU2892`
    double areatriangle(Point a, Point b)
    {
        if (sgn((p - a) ^ (p - b)) == 0) return 0.0;
        Point q[5],p1, p2;
        int len = 0;
        q[len++] = a;
        Line l(a, b);
        if (pointcrossline(l, q[1], q[2]) == 2)
        {
            if (sgn((a - q[1]) * (b - q[1])) < 0) q[len++] = q[1];
            if (sgn((a - q[2]) * (b - q[2])) < 0) q[len++] = q[2];
        }
        q[len++] = b;
        if (len == 4 && sgn((q[0] - q[1]) * (q[2] - q[1])) > 0) swap(q[1], q[2]);
        double res = 0;
        for (int i = 0; i < len - 1; i++)
        {
            if (relation(q[i]) == 0 || relation(q[i + 1]) == 0)
            {
                double arg = p.rad(q[i], q[i + 1]);
                res += r * r * arg / 2.0;
            }
            else res += fabs((q[i] - p) ^ (q[i + 1] - p)) / 2.0;
        }
        return res;
    }
    //`两圆公切线`
    Point getpoint(double rad) { return Point(p.x + r * cos(rad), p.y + r * sin(rad)); }
    int conmontangent(circle v, vector<Point> &p1, vector<Point> &p2)
    {
        bool flag = 0;
        if (r < v.r) swap(*this, v), flag = 1;
        double d = p.distance(v.p), rd = r - v.r, rs = r + v.r;
        if (sgn(d - rd) < 0) return 0;
        if (sgn(d) == 0) return -1;
        double rad = Line(p, v.p).angle();
        if (sgn(d - rd) == 0)
        {
            p1.push_back(getpoint(rad)), p2.push_back(getpoint(rad));
            return 1; //`一条外公切线`
        }
        double rad1 = acos(rd / d);
        p1.push_back(getpoint(rad + rad1)), p2.push_back(v.getpoint(rad + rad1));
        p1.push_back(getpoint(rad - rad1)), p2.push_back(v.getpoint(rad - rad1));
        if (sgn(d - rs) == 0)
        {
            p1.push_back(getpoint(rad)), p2.push_back(getpoint(rad));
            if (flag) swap(p1, p2);
            return 3; //`两条外公切线 一条内公切线`
        }
        else if (sgn(d - rs) > 0)
        {
            double rad2 = acos(rs / d);
            p1.push_back(getpoint(rad + rad2)), p2.push_back(v.getpoint(rad + rad2 - pi));
            p1.push_back(getpoint(rad - rad2)), p2.push_back(v.getpoint(rad - rad2 + pi));
            if (flag) swap(p1, p2);
            return 4; //`两条外公切线 两条内公切线`
        }
        else
        {
            if (flag) swap(p1, p2);
            return 2; //`两条外公切线`
        }
    }
};

struct polygon
{
    int n;
    Point p[N];
    Line l[N];
    void input(int _n)
    {
        n = _n;
        for (int i = 0; i < n; i++) p[i].input();
    }
    void add(Point q) { p[n++] = q; }
    void getline()
    {
        for (int i = 0; i < n; i++)
            l[i] = Line(p[i], p[(i + 1) % n]);
    }
    struct cmp
    {
        Point p;
        cmp(const Point &p0) { p = p0; }
        bool operator()(const Point &aa, const Point &bb)
        {
            Point a = aa, b = bb;
            int d = sgn((a - p) ^ (b - p));
            if (d == 0) return sgn(a.distance(p) - b.distance(p)) < 0;
            return d > 0;
        }
    };
    //`进行极角排序 mi为极点`
    //`需要重载号好Point的 < 操作符(min函数要用) `
    void norm(Point mi)
    {
        // Point mi = p[0];
        // for(int i = 1;i < n;i++)mi = min(mi,p[i]);
        sort(p, p + n, cmp(mi));
    }
    //`得到的凸包里面的点编号是0-n-1的`
    //`注意如果有影响，要特判下所有点共点，或者共线的特殊情况`
    //`测试 LightOJ1203  LightOJ1239`
    void andrew(polygon &convex)
    {
        sort(p, p + n);
        int &top = convex.n;
        top = 0;
        for (int i = 0; i < n; i++)
        {
            while (top >= 2 && sgn(cross(convex.p[top - 2], convex.p[top - 1], p[i])) <= 0) top--;
            convex.p[top++] = p[i];
        }
        int temp = top;
        for (int i = n - 2; i >= 0; i--)
        {
            while (top > temp && sgn(cross(convex.p[top - 2], convex.p[top - 1], p[i])) <= 0) top--;
            convex.p[top++] = p[i];
        }
        top--;
    }
    //`判断是不是凸的`
    bool isconvex()
    {
        bool s[5];
        memset(s, false, sizeof(s));
        for (int i = 0; i < n; i++)
        {
            int j = (i + 1) % n, k = (j + 1) % n;
            s[sgn((p[j] - p[i]) ^ (p[k] - p[i])) + 1] = true;
            if (s[0] && s[2]) return false;
        }
        return true;
    }
    //`判断点和任意多边形的关系`
    int relationpoint(Point q)
    {
        for (int i = 0; i < n; i++)
            if (p[i] == q) return 3;                 //` 3 点上`
        for (int i = 0; i < n; i++)
            if (l[i].pointonseg(q)) return 2;        //` 2 边上`
        int cnt = 0;
        for (int i = 0; i < n; i++)
        {
            int j = (i + 1) % n;
            int k = sgn((q - p[j]) ^ (p[i] - p[j]));
            int u = sgn(p[i].y - q.y);
            int v = sgn(p[j].y - q.y);
            if (k > 0 && u < 0 && v >= 0) cnt++;
            if (k < 0 && v < 0 && u >= 0) cnt--;
        }                                            //` 1 内部`
        return cnt != 0;                             //` 0 外部`
    }
    //`直线u切割凸多边形左侧 注意直线方向`
    //`测试：HDU3982`
    void convexcut(Line u, polygon &po)
    {
        int &top = po.n; // 注意引用
        top = 0;
        for (int i = 0; i < n; i++)
        {
            int d1 = sgn((u.e - u.s) ^ (p[i] - u.s));
            int d2 = sgn((u.e - u.s) ^ (p[(i + 1) % n] - u.s));
            if (d1 >= 0) po.p[top++] = p[i];
            if (d1 * d2 < 0) po.p[top++] = u.crosspoint(Line(p[i], p[(i + 1) % n]));
        }
    }
    //`得到周长`
    //`测试 LightOJ1239`
    double getcircumference()
    {
        double sum = 0;
        for (int i = 0; i < n; i++)
            sum += p[i].distance(p[(i + 1) % n]);
        return sum;
    }
    //`得到面积`
    double getarea()
    {
        double sum = 0;
        for (int i = 0; i < n; i++)
            sum += (p[i] ^ p[(i + 1) % n]);
        return fabs(sum) / 2;
    }
    //`得到方向`
    bool getdir()
    {
        double sum = 0;
        for (int i = 0; i < n; i++)
            sum += (p[i] ^ p[(i + 1) % n]);
        if (sgn(sum) > 0) return 1; //` 1 逆时针`
        else return 0;              //` 0 顺时针`
    }
    //`得到重心`
    Point getbarycentre()
    {
        Point ret(0, 0);
        double area = 0;
        for (int i = 1; i < n - 1; i++)
        {
            double tmp = (p[i] - p[0]) ^ (p[i + 1] - p[0]);
            if (sgn(tmp) == 0) continue;
            area += tmp;
            ret.x += (p[0].x + p[i].x + p[i + 1].x) / 3 * tmp;
            ret.y += (p[0].y + p[i].y + p[i + 1].y) / 3 * tmp;
        }
        if (sgn(area)) ret = ret / area;
        return ret;
    }
    //`多边形和圆交的面积`
    //`测试：POJ3675 HDU3982 HDU2892`
    double areacircle(circle c)
    {
        double ans = 0;
        for (int i = 0; i < n; i++)
        {
            int j = (i + 1) % n;
            if (sgn((p[j] - c.p) ^ (p[i] - c.p)) >= 0) ans += c.areatriangle(p[i], p[j]);
            else ans -= c.areatriangle(p[i], p[j]);
        }
        return fabs(ans);
    }
    //`多边形和圆关系`
    int relationcircle(circle c)
    {
        int x = 2;                                  //` 2 圆完全在多边形内`
        if (relationpoint(c.p) != 1) return 0;      //` 0 圆心不在内部`
        for (int i = 0; i < n; i++)
        {
            if (c.relationseg(l[i]) == 2) return 0; //` 0 其它`
            if (c.relationseg(l[i]) == 1) x = 1;    //` 1 圆在多边形里面，碰到了多边形边界`
        }
        return x;
    }
    //`旋转卡壳求凸包直径(最远点对)`
    double rorating_calipers1()
    {
        double res = 0;
        for (int i = 0, j = 1; i < n; i++)
        {
            while (dsgn(cross(p[i], p[i + 1], p[j]), cross(p[i], p[i + 1], p[j + 1])) < 0) j = (j + 1) % n;
            res = max(res, max(p[i].distance(p[j]), p[i + 1].distance(p[j])));
        }
        return res;
    }
    //`旋转卡壳求最小矩形覆盖`
    double rorating_calipers2(polygon &pt)
    {
        double res = 1e20;
        for (int i = 0, a = 1, b = 1, c; i < n; i++)
        {
            while (dsgn(cross(p[i], p[i + 1], p[a]), cross(p[i], p[i + 1], p[a + 1])) < 0) a = (a + 1) % n;
            while (dsgn(dot(p[i], p[i + 1], p[b]), dot(p[i], p[i + 1], p[b + 1])) < 0) b = (b + 1) % n;
            if (!i) c = a;
            while (dsgn(dot(p[i + 1], p[i], p[c]), dot(p[i + 1], p[i], p[c + 1])) < 0) c = (c + 1) % n;
            double d = p[i].distance(p[i + 1]);
            double H = cross(p[i], p[i + 1], p[a]) / d;
            double R = dot(p[i], p[i + 1], p[b]) / d;
            double L = dot(p[i + 1], p[i], p[c]) / d;
            if (dsgn(res, (L + R - d) * H) > 0)
            {
                res = (L + R - d) * H;
                pt.p[0] = p[i + 1] + (p[i] - p[i + 1]) * (L / d);
                pt.p[1] = p[i] + (p[i + 1] - p[i]) * (R / d);
                pt.p[2] = pt.p[1] + (p[i + 1] - p[i]).rotleft() * (H / d);
                pt.p[3] = pt.p[0] + (p[i + 1] - p[i]).rotleft() * (H / d);
            }
        }
        return res;
    }
    //`分治法求最近点对`
    Point a[N];
    double divide(int l, int r)
    {
        if (l == r) return 2e9;
        if (l + 1 == r) return p[l].distance(p[r]);
        int mid = l + r >> 1;
        double d = min(divide(l, mid), divide(mid + 1, r));
        int k = 0;
        for (int i = l; i <= r; i++)
            if (fabs(p[mid].x - p[i].x) < d) a[k++] = p[i];
        // sort(a,a+k,[&](Point a,Point b)->bool {return a.y<b.y;});
        for (int i = 0; i < k; i++)
            for (int j = i + 1; j < k && a[j].y - a[i].y < d; j++)
                d = min(d, a[j].distance(a[i]));
        return d;
    }
    //`旋转卡壳求最大三角形面积`
    double rotating_calipers3()
    {
        double res = 0;
        for (int i = 0; i < n; i++)
        {
            int k = i + 1;
            for (int j = i + 1; j < n; j++)
            {
                while (dsgn(cross(p[i], p[j], p[k]), cross(p[i], p[j], p[k + 1])) < 0) k = (k + 1) % n;
                res = max(res, cross(p[i], p[j], p[k]));
            }
        }
        return res / 2;
    }
};
//`半平面交求凸多边形面积交`
double half_plane1(Line l[], int n)
{
    double res = 0;
    sort(l, l + n);
    Line q[N];
    Point p[N];
    int h = 0, t = 0, k = 0;
    q[t++] = l[0];
    for (int i = 1; i < n; i++)
    {
        if (sgn(l[i].angle() - l[i - 1].angle()) == 0) continue;
        while (h < t - 1 && l[i].relation(q[t - 1].crosspoint(q[t - 2])) == 2) t--;
        while (h < t - 1 && l[i].relation(q[h].crosspoint(q[h + 1])) == 2) h++;
        q[t++] = l[i];
    }
    while (h < t - 1 && l[h].relation(q[t - 1].crosspoint(q[t - 2])) == 2) t--;
    q[t++] = q[h];
    for (int i = h; i < t - 1; i++) p[k++] = q[i].crosspoint(q[i + 1]);
    for (int i = 1; i < k - 1; i++) res += (p[i] - p[0]) ^ (p[i + 1] - p[0]);
    return res / 2;
}
//`水平可见直线 从上向下看输出能看见哪些直线`
void half_plane2(Line l[], int n)
{
    sort(l, l + n);
    Line q[N];
    int h = 0, t = 0, k = 0;
    q[t++] = l[0];
    for (int i = 1; i < n; i++)
    {
        if (sgn(l[i].angle() - l[i - 1].angle()) == 0) continue;
        while (h < t - 1 && l[i].relation(q[t - 1].crosspoint(q[t - 2])) == 2) t--;
        // while(h<t-1&&l[i].relation(q[h].crosspoint(q[h+1]))==2) h++;
        q[t++] = l[i];
    }
    int ans[N];
    for (int i = h; i < t; i++)
        // for(auto j:q[i].id) ans[k++]=j;
    sort(ans, ans + k);
    cout << k << endl;
    for (int i = 0; i < k; i++) cout << ans[i] << " ";
}
//`多边形内核`
bool half_plane3(Line l[], int n)
{
    sort(l, l + n);
    Line q[N];
    int h = 0, t = 0;
    q[t++] = l[0];
    for (int i = 1; i < n; i++)
    {
        if (sgn(l[i].angle() - l[i - 1].angle()) == 0) continue;
        while (h < t - 1 && l[i].relation(q[t - 1].crosspoint(q[t - 2])) == 2) t--;
        while (h < t - 1 && l[i].relation(q[h].crosspoint(q[h + 1])) == 2) h++;
        q[t++] = l[i];
    }
    while (h < t - 1 && l[h].relation(q[t - 1].crosspoint(q[t - 2])) == 2) t--;
    return t - h >= 3;
}
//`最小圆覆盖`
circle increment(Point p[], int n)
{
    random_shuffle(p, p + n);
    circle ans;
    ans.p = p[0], ans.r = 0;
    for (int i = 1; i < n; i++)
        if (ans.r < ans.p.distance(p[i]))
        {
            ans.p = p[i], ans.r = 0;
            for (int j = 0; j < i; j++)
                if (ans.r < ans.p.distance(p[j]))
                {
                    ans.p = (p[i] + p[j]) / 2, ans.r = p[i].distance(p[j]) / 2;
                    for (int k = 0; k < j; k++)
                        if (ans.r < ans.p.distance(p[k]))
                        {
                            Point p1 = (p[i] + p[j]) / 2;
                            Point v1 = (p[i] - p[j]).rotright();
                            Point p2 = (p[i] + p[k]) / 2;
                            Point v2 = (p[i] - p[k]).rotright();
                            ans.p = Line(p1, p1 + v1).crosspoint(Line(p2, p2 + v2));
                            ans.r = ans.p.distance(p[i]);
                        }
                }
        }
    return ans;
}
//`自适应辛普森积分`
inline double f(double x)
{ // 积分函数
    return x;
}
double simpson(double l, double r)
{ // 辛普森公式
    double mid = (l + r) / 2;
    return (r - l) * (f(l) + f(r) + 4 * f(mid)) / 6;
}
double asr(double l, double r, double ans)
{ // 自适应
    double mid = (l + r) / 2;
    double a = simpson(l, mid), b = simpson(mid, r);
    if (sgn(a + b - ans) == 0) return ans;
    return asr(l, mid, a) + asr(mid, r, b);
}

struct Point3
{
    double x, y, z;
    //Point3(){}
    Point3(double _x=0, double _y=0, double _z=0) { x = _x;y = _y;z = _z; }
    void input() { cin>>x>>y>>z; }
    bool operator==(const Point3 &b) const {return sgn(x-b.x)==0&&sgn(y-b.y)==0&&sgn(z-b.z)==0;}
    bool operator<(const Point3 &b) const {return sgn(x-b.x)==0?(sgn(y-b.y)==0?sgn(z-b.z)<0:y<b.y):x<b.x;}
    double len() { return sqrt(x * x + y * y + z * z); }
    double len2() { return x * x + y * y + z * z; }
    double distance(const Point3 &b) const {return sqrt((x-b.x)*(x-b.x)+(y-b.y)*(y-b.y)+(z-b.z)*(z-b.z));}
    Point3 operator-(const Point3 &b) const { return Point3(x - b.x, y - b.y, z - b.z); }
    Point3 operator+(const Point3 &b) const { return Point3(x + b.x, y + b.y, z + b.z); }
    Point3 operator*(const double &k) const { return Point3(x * k, y * k, z * k); }
    Point3 operator/(const double &k) const { return Point3(x / k, y / k, z / k); }
    // 点乘
    double operator*(const Point3 &b) const { return x * b.x + y * b.y + z * b.z; }
    // 叉乘
    Point3 operator^(const Point3 &b) const {return Point3(y*b.z-z*b.y,z*b.x-x*b.z,x*b.y-y*b.x);}
    double rad(Point3 a, Point3 b)
    {
        Point3 p = (*this);
        return acos(((a - p) * (b - p)) / (a.distance(p) * b.distance(p)));
    }
    //`化为长度为r的向量`
    Point3 trunc(double r)
    {
        double l = len();
        if (!sgn(l)) return *this;
        r /= l;
        return Point3(x * r, y * r, z * r);
    }
};

struct Line3
{
    Point3 s, e;
    Line3() {}
    Line3(Point3 _s, Point3 _e) { s = _s;e = _e; }
    bool operator==(const Line3 v) { return (s == v.s) && (e == v.e); }
    void input() { s.input();e.input(); }
    double length() { return s.distance(e); }
    // 点到直线距离
    double dispointtoline(Point3 p) { return ((e - s) ^ (p - s)).len() / s.distance(e); }
    // 点到线段距离
    double dispointtoseg(Point3 p)
    {
        if (sgn((p - s) * (e - s)) < 0 || sgn((p - e) * (s - e)) < 0) return min(p.distance(s), e.distance(p));
        return dispointtoline(p);
    }
    //`返回点p在直线上的投影`
    Point3 lineprog(Point3 p) { return s + (((e - s) * ((e - s) * (p - s))) / ((e - s).len2())); }
    //`p绕此向量逆时针arg角度`
    Point3 rotate(Point3 p, double ang)
    {
        if (sgn(((s - p) ^ (e - p)).len()) == 0) return p;
        Point3 f1 = (e - s) ^ (p - s);
        Point3 f2 = (e - s) ^ (f1);
        double len = ((s - p) ^ (e - p)).len() / s.distance(e);
        f1 = f1.trunc(len);
        f2 = f2.trunc(len);
        Point3 h = p + f2;
        Point3 pp = h + f1;
        return h + ((p - h) * cos(ang)) + ((pp - h) * sin(ang));
    }
    //`点在直线上`
    bool pointonseg(Point3 p) { return sgn(((s - p) ^ (e - p)).len()) == 0 && sgn((s - p) * (e - p)) == 0; }
};

struct Plane
{
    Point3 a, b, c, o; //`平面上的三个点，以及法向量`
    Plane() {}
    Plane(Point3 _a, Point3 _b, Point3 _c)
    {
        a = _a;b = _b;c = _c;
        o = pvec();
    }
    Point3 pvec() { return (b - a) ^ (c - a); }
    //`ax+by+cz+d = 0`
    Plane(double _a, double _b, double _c, double _d)
    {
        o = Point3(_a, _b, _c);
        if (sgn(_a) != 0) a = Point3((-_d - _c - _b) / _a, 1, 1);
        else if (sgn(_b) != 0) a = Point3(1, (-_d - _c - _a) / _b, 1);
        else if (sgn(_c) != 0) a = Point3(1, 1, (-_d - _a - _b) / _c);
    }
    //`点在平面上的判断`
    bool pointonplane(Point3 p) { return sgn((p - a) * o) == 0; }
    //`两平面夹角`
    double angleplane(Plane f) { return acos(o * f.o) / (o.len() * f.o.len()); }
    //`平面和直线的交点，返回值是交点个数`
    int crossline(Line3 u, Point3 &p)
    {
        double x = o * (u.e - a);
        double y = o * (u.s - a);
        double d = x - y;
        if (sgn(d) == 0) return 0;
        p = ((u.s * x) - (u.e * y)) / d;
        return 1;
    }
    //`点到平面最近点(也就是投影)`
    Point3 pointtoplane(Point3 p)
    {
        Line3 u = Line3(p, p + o);
        crossline(u, p);
        return p;
    }
    //`平面和平面的交线`
    int crossplane(Plane f, Line3 &u)
    {
        Point3 oo = o ^ f.o;
        Point3 v = o ^ oo;
        double d = fabs(f.o * v);
        if (sgn(d) == 0) return 0;
        Point3 q = a + (v * (f.o * (f.a - a)) / d);
        u = Line3(q, q + oo);
        return 1;
    }
};

struct CH3D
{
    struct face
    {
        int a, b, c; // 表示凸包一个面上的三个点的编号
        bool ok;     // 表示该面是否属于最终的凸包上的面
    };
    int n;// 初始顶点数
    Point3 P[N];
    int num;// 凸包表面的三角形数
    face F[8 * N];// 凸包表面的三角形
    int g[N][N];
    void input(int _n)
    {
        n = _n;
        for (int i = 0; i < n; i++) P[i].input();
    }
    // 叉乘
    Point3 cross(const Point3 &a, const Point3 &b, const Point3 &c) { return (b - a) ^ (c - a); }
    //`三角形面积*2`
    double area(Point3 a, Point3 b, Point3 c) { return ((b - a) ^ (c - a)).len(); }
    //`四面体有向面积*6`
    double volume(Point3 a, Point3 b, Point3 c, Point3 d) { return ((b - a) ^ (c - a)) * (d - a); }
    //`正：点在面同向`
    double dblcmp(Point3 &p, face &f)
    {
        Point3 p1 = P[f.b] - P[f.a];
        Point3 p2 = P[f.c] - P[f.a];
        Point3 p3 = p - P[f.a];
        return (p1 ^ p2) * p3;
    }
    void deal(int p, int a, int b)
    {
        int f = g[a][b];
        face add;
        if (F[f].ok)
        {
            if (dblcmp(P[p], F[f]) > eps) dfs(p, f);
            else
            {
                add.a = b;
                add.b = a;
                add.c = p;
                add.ok = true;
                g[p][b] = g[a][p] = g[b][a] = num;
                F[num++] = add;
            }
        }
    }
    // 递归搜索所有应该从凸包内删除的面
    void dfs(int p, int now)
    {
        F[now].ok = false;
        deal(p, F[now].b, F[now].a);
        deal(p, F[now].c, F[now].b);
        deal(p, F[now].a, F[now].c);
    }
    bool same(int s, int t)
    {
        Point3 &a = P[F[s].a];
        Point3 &b = P[F[s].b];
        Point3 &c = P[F[s].c];
        return fabs(volume(a, b, c, P[F[t].a])) < eps &&
               fabs(volume(a, b, c, P[F[t].b])) < eps &&
               fabs(volume(a, b, c, P[F[t].c])) < eps;
    }
    // 构建三维凸包
    void create()
    {
        num = 0;
        face add;
        //***********************************
        // 此段是为了保证前四个点不共面
        bool flag = true;
        for (int i = 1; i < n; i++)
        {
            if (!(P[0] == P[i]))
            {
                swap(P[1], P[i]);
                flag = false;
                break;
            }
        }
        if (flag) return;
        flag = true;
        for (int i = 2; i < n; i++)
        {
            if (((P[1] - P[0]) ^ (P[i] - P[0])).len() > eps)
            {
                swap(P[2], P[i]);
                flag = false;
                break;
            }
        }
        if (flag)  return;
        flag = true;
        for (int i = 3; i < n; i++)
        {
            if (fabs(((P[1] - P[0]) ^ (P[2] - P[0])) * (P[i] - P[0])) > eps)
            {
                swap(P[3], P[i]);
                flag = false;
                break;
            }
        }
        if (flag) return;
        //**********************************
        for (int i = 0; i < 4; i++)
        {
            add.a = (i + 1) % 4;
            add.b = (i + 2) % 4;
            add.c = (i + 3) % 4;
            add.ok = true;
            if (dblcmp(P[i], add) > 0) swap(add.b, add.c);
            g[add.a][add.b] = g[add.b][add.c] = g[add.c][add.a] = num;
            F[num++] = add;
        }
        for (int i = 4; i < n; i++)
            for (int j = 0; j < num; j++)
                if (F[j].ok && dblcmp(P[i], F[j]) > eps)
                {
                    dfs(i, j);
                    break;
                }
        int tmp = num;
        num = 0;
        for (int i = 0; i < tmp; i++)
            if (F[i].ok) F[num++] = F[i];
    }
    // 表面积
    //`测试：HDU3528`
    double area()
    {
        double res = 0;
        if (n == 3)
        {
            Point3 p = cross(P[0], P[1], P[2]);
            return p.len() / 2;
        }
        for (int i = 0; i < num; i++)
            res += area(P[F[i].a], P[F[i].b], P[F[i].c]);
        return res / 2.0;
    }
    double volume()
    {
        double res = 0;
        Point3 tmp = Point3(0, 0, 0);
        for (int i = 0; i < num; i++)
            res += volume(tmp, P[F[i].a], P[F[i].b], P[F[i].c]);
        return fabs(res / 6);
    }
    // 表面三角形个数
    int triangle() { return num; }
    // 表面多边形个数
    //`测试：HDU3662`
    int polygon()
    {
        int res = 0;
        for (int i = 0; i < num; i++)
        {
            bool flag = true;
            for (int j = 0; j < i; j++)
                if (same(i, j))
                {
                    flag = 0;
                    break;
                }
            res += flag;
        }
        return res;
    }
    // 重心
    //`测试：HDU4273`
    Point3 barycenter()
    {
        Point3 ans = Point3(0, 0, 0);
        Point3 o = Point3(0, 0, 0);
        double all = 0;
        for (int i = 0; i < num; i++)
        {
            double vol = volume(o, P[F[i].a], P[F[i].b], P[F[i].c]);
            ans = ans + (((o + P[F[i].a] + P[F[i].b] + P[F[i].c]) / 4.0) * vol);
            all += vol;
        }
        ans = ans / all;
        return ans;
    }
    // 点到面的距离
    //`测试：HDU4273`
    double ptoface(Point3 p, int i)
    {
        double tmp1 = fabs(volume(P[F[i].a], P[F[i].b], P[F[i].c], p));
        double tmp2 = ((P[F[i].b] - P[F[i].a]) ^ (P[F[i].c] - P[F[i].a])).len();
        return tmp1 / tmp2;
    }
};

template<typename T>
inline T read()
{
    T x = 0;int f = 1;
    char ch = getchar();
    while (ch < '0' || ch > '9')
    {
        if (ch == '-') f = -1;
        ch = getchar();
    }
    while (ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar();
    return x * f;
}
void solve()
{
    double l1,l2,a1,a2;cin>>l1>>l2>>a1>>a2;
    double A=a1/180*pi,B=a2/180*pi;
    double s1=A*(l1+l2)*(l1+l2),s2=B*l2*l2;
    if(a2<=90) cout<<fixed<<setprecision(10)<<s1+s2<<endl;
    else
    {
        double l3=sqrt(l1*l1+l2*l2-2*l1*l2*cos(pi-B));
        double C=asin(l2/l3*sin(B)),D=B-C;      
        if(D<=pi/2&&B>=pi/2&&C<=pi/2)
        {
            double h=l1*l3*sin(C)/l2,s3;
            if(B-pi/2>=2*A) s3=h*l1*sin(B-pi/2)/2-h*h*A-h*h*tan(B-pi/2-2*A)/2;
            else s3=h*l1*sin(B-pi/2)/2-h*h*(B-pi/2)/2;
            cout<<fixed<<setprecision(10)<<s1+s2+2*s3<<endl;
        }
        else
        {
            if(C<=2*A)
            {
                double s3=l1*l3*sin(C)-C*l3*l3;
                cout<<fixed<<setprecision(10)<<s1+s2+s3<<endl;
            }
            else
            {
                double s3=l1*l3*sin(C)/2-A*l3*l3-sin(C-2*A)*l3*l3*sin(D)/sin(pi-D-(C-2*A))/2;
                cout<<fixed<<setprecision(10)<<s1+s2+2*s3<<endl;
            }
        }
    }
}

int main()
{
    //ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
    int t = 1; cin >> t;
    while (t--) solve();
    return 0;
}