#include <bits/stdc++.h>
#define pi acos(-1.0)
//#define double long double
using namespace std;
const double eps = 1e-8,inf=1e12;
const int N=1e3+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);}
    int 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];
        int len = 0;
        q[len++] = a;
        Line l(a,b);
        Point p1,p2;
        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));
    }

    //`得到凸包`
    //`得到的凸包里面的点编号是0-n-1的`
    //`两种凸包的方法`
    //`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`
        }
        getline();
        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){
        getline();
        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);
}

void solve(int cnt)
{
    int n;double r;cin>>n>>r;
    circle o=circle(Point(0,0),r);
    circle c[N];
    polygon p;p.n=0;
    for(int i=0;i<n;i++)
    {
        double x,y,R;cin>>x>>y>>R;
        c[i]=circle(Point(x,y),R);
        Point p1,p2;
        int t=o.pointcrosscircle(c[i],p1,p2);
        if(t==2) p.add(p1),p.add(p2);
        else if(t==1) p.add(p1);
    }
    for(int i=0;i<p.n;i++)
    {
        bool f=0;
        Point t=Point(-p.p[i].x,-p.p[i].y);
        for(int j=0;j<n;j++)
            if(c[j].relation(t))
            {
                f=1;
                break;
            }
        if(!f)
        {
            cout<<"Case #"<<cnt<<": "<<fixed<<setprecision(10)<<r<<endl;
            return;
        }      
    }   
    polygon t;
    p.andrew(t);
    cout<<"Case #"<<cnt<<": "<<fixed<<setprecision(10)<<t.rorating_calipers1()<<endl;
}

int main()
{
    ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
    int t=1;cin>>t;
    for(int i=1;i<=t;i++) solve(i);
    return 0;
}