#include <bits/stdc++.h>
using namespace std;
#define endl '\n'
#define IOS ios::sync_with_stdio(0); cin.tie(0); cout.tie(0)
#define pb push_back

#define Yes cout << "Yes\n"
#define No cout << "No\n"
#define YES cout << "YES\n"
#define NO cout << "NO\n"
#define ff first
#define ss second
#define fori(x,y) for(int i=x;i<=(int)(y);++i)
#define forj(x,y) for(int j=x;j<=(int)(y);++j)
#define fork(x,y) for(int k=x;k<=(int)(y);++k)

#define debug(x) cerr << #x << " = " << x << endl

typedef long long ll;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;

ll MOD = 998244353;
ll qpow(ll a,ll p) {ll res=1; while(p) {if (p&1) {res=res*a%MOD;} a=a*a%MOD; p>>=1;} return res;}
namespace Geometry {
    const double PI = acos(-1);
    const double eps = 1e-13;
    int cmp(double x, double y) {  // 比较两个浮点数的大小 x<y为-1，x>y为1
        if (fabs(x-y)<eps) return 0;
        else return x<y?-1:1;
    }
    int sgn(double x) {  // 符号函数，大于0为1，小于0为-1
        return cmp(x, 0);
    }

    struct Vector {
        double x, y;
        Vector(): x(0), y(0) {}
        Vector(double x, double y): x(x), y(y) {}
        
        bool zero() const {
            return sgn(x) == 0&&sgn(y) == 0;
        }
        double distance2(const Vector& B) const {
            return (x-B.x)*(x-B.x)+(y-B.y)*(y-B.y);
        }
        double distance(const Vector& B) const {
            return sqrt(distance2(B));
        }
        double length2() const {
            return x*x+y*y;
        }
        double length() const {
            return sqrt(length2());
        }
        Vector normalize() const {
            return Vector(x/length(), y/length());
        }

        Vector operator+(const Vector& B) const {
            return Vector(x+B.x, y+B.y);
        }
        Vector operator-(const Vector& B) const {
            return Vector(x-B.x, y-B.y);
        }
        Vector operator*(double k) const {
            return Vector(x*k, y*k);
        }
        Vector operator/(double k) const {
            return Vector(x/k, y/k);
        }
        double dot(const Vector& B) const {
            return x*B.x+y*B.y;
        }
        double operator*(const Vector& B) const {
            return dot(B);
        }
        bool operator==(const Vector& B) const {
            return sgn(x-B.x) == 0&&sgn(y-B.y) == 0;
        }
        double cross(const Vector& B) const {  // 从自身旋转到B的叉积（注意方向，逆时针正顺时针负）
            return x*B.y-y*B.x;
        }
        double angle(const Vector& B) const {  // 计算与另一个向量的夹角（注意方向，逆时针正顺时针负）
            return sgn(cross(B))*acos(dot(B)/length()/B.length());
        }
        bool parallel(const Vector& B) const {  // 判断两个向量是否平行
            return sgn(cross(B)) == 0;
        }
    };

    struct Polygon {  // 由向量构成的多边形
        vector<Vector> points;  // 尽量用逆时针排布
        Polygon(): points(vector<Vector>()) {}
        Polygon(vector<Vector> points): points(points) {}
        
        /**
         * 判断点是否在多边形内：
         * 3：点在多边形顶点上
         * 2：点在多边形的边上
         * 1：点在多边形内部
         * 0：点在多边形外部
         */
        int isInPolygon(Vector point) const {
            int n = points.size();
            int num = 0;
            fori(0, n-1) {
                Vector v1 = points[i], v2 = points[(i+1)%n];
                int c = sgn((point-v2).cross(v1-v2));
                int u = sgn(v1.y-point.y);
                int v = sgn(v2.y-point.y);
                if (c>0&&u<0&&v>=0) num++;
                if (c<0&&u>=0&&v<0) num--;
            }
            return num!=0;
        }
        /**
         * 求多边形面积的两倍
         * 但一定要注意！多边形得是逆时针排布！
         */
        double area2() const {
            double sum = 0;
            int n = points.size();
            fori(0, n-1) sum += points[i].cross(points[(i+1)%n]);
            return sum;
        }
        double area() const {  // 多边形面积
            return area2()/2;
        }

        /**
         * 求多边形的重心
         */
        Vector gravityCenter() const {
            Vector ans;
            int n = points.size();
            fori(0, n-1) {
                Vector v1 = points[i], v2 = points[(i+1)%n];
                ans = ans+(v1+v2)*v1.cross(v2);
            }
            ans = ans/area()/6;
            return ans;
        }
    };

    struct Line {  // 直线
        Vector p1, p2;  // 两点确定一条直线
        Line(Vector p1, Vector p2): p1(p1), p2(p2) {}  // 基础构造方法
        Line(Vector p, double angle) {  // 斜率构造方法：一点+斜率，0<=angle<=pi
            p1 = p;
            if (sgn(angle-PI/2) == 0) {p2 = (p1+Vector(0, 1));}
            else {p2 = (p1+Vector(1, tan(angle)));}
        }
        Line(double a, double b, double c) {  // ax+by+c=0
            if (sgn(a) == 0) {
                p1 = Vector(0, -c/b);
                p2 = Vector(1, -c/b);
            }
            else if (sgn(b) == 0) {
                p1 = Vector(-c/a, 0);
                p2 = Vector(-c/a, 1);
            }
            else {
                p1 = Vector(0, -c/b);
                p2 = Vector(1, -(c+a)/b);
            }
        }

        // 判断一个点在直线上、左侧还是右侧
        int pointLineRelation(Vector p) const {
            int c = sgn((p-p1).cross(p2-p1));
            if (c<0) return 1;  // 1: 左侧
            else if (c>0) return 2;  // 2: 右侧
            else return 0;  // 0: 在直线上
        }

        // 计算点到直线的距离
        double distanceToPoint(Vector p) const {
            return fabs((p-p1).cross(p2-p1))/p1.distance(p2);
        }

        // 点在直线上的投影
        Vector pointProj(Vector p) const {
            double k = (p-p1).dot(p2-p1)/p1.distance2(p2);
            return p1+(p2-p1)*k;
        }

        // 与另一条直线的位置关系
        int lineRelation(const Line& B) const {
            if (sgn((p2-p1).cross(B.p2-B.p1)) == 0) {
                if (pointLineRelation(B.p1) == 0) return 1;  // 1: 重合
                else return 0;  // 0: 平行
            }
            return 2;  // 2: 相交
        }
        
        // 与另一条直线的交点（注意！两条直线必须相交！！！）
        Vector crossPoint(const Line& B) const {
            double s1 = (p2-p1).cross(B.p1-p1);
            double s2 = (p2-p1).cross(B.p2-p1);
            return Vector(B.p1.x*s2-B.p2.x*s1, B.p1.y*s2-B.p2.y*s1)/(s2-s1);
        }
    };

    struct Circle {
        Vector c;  // 圆心
        double r;  // 半径
        Circle(Vector c, double r): c(c), r(r) {}
        Circle(const Vector& a, const Vector& b, const Vector& c) {  // 三点定圆
            double a1 = b.x-a.x, b1 = b.y-a.y, c1 = (a1*a1+b1*b1)/2;
            double a2 = c.x-a.x, b2 = c.y-a.y, c2 = (a2*a2+b2*b2)/2;
            double d = a1*b2-a2*b1;
            Vector center(a.x+(c1*b2-c2*b1)/d, a.y+(a1*c2-a2*c1)/d);
            this->c = center;
            r = center.distance(a);
        }

        // 点和圆的位置关系
        int pointCircleRelation(Vector p) const {
            double d = p.distance(c);
            if (sgn(d-r)<0) return 0;  // 0: 点在圆内
            else if (sgn(d-r)==0) return 1;  // 1: 点在圆上
            else return 2;  // 2: 点在圆外
        }

        // 直线和圆的位置关系
        int lineCircleRelation(const Line& line) const {
            double d = line.distanceToPoint(c);
            if (sgn(d-r)<0) return 0;  // 0: 直线和圆相交
            else if (sgn(d-r)==0) return 1;  // 1: 直线和圆相切
            else return 2;  // 2: 直线在圆外
        }

        /**
         * 求解直线和圆的交点
         * 
         * @param line 直线
         * @param p1 返回第一个交点
         * @param p2 返回第二个交点
         * @return 返回交点的个数
         */ 
        int lineCrossCircle(const Line& line, Vector& p1, Vector& p2) const {
            if (lineCircleRelation(line) == 2) return 0;  // 没有交点
            Vector pro = line.pointProj(c);
            double d = line.distanceToPoint(c);
            double k = sqrt(r*r-d*d);
            if (sgn(k) == 0) {
                p1 = pro, p2 = pro;
                return 1;  // 直线和圆相切，只有一个交点
            }
            Vector n = (line.p2-line.p1)/line.p2.distance(line.p1);
            p1 = pro+n*k, p2 = pro-n*k;
            return 2;  // 两个交点
        }
    };
};
using namespace Geometry;

// Andrew算法求凸包
// 先求下凸包再求上凸包
// sets是按照x从小到大，y从小到大排序的点对
Polygon getConvexHull(const set<pair<double,double>>& sets) {
    // 下凸包
    vector<Vector> down;
    for (auto it=sets.begin();it!=sets.end();++it) {  // 从前到后遍历sets
        Vector p(it->first, it->second);
        while (down.size()>1) {
            Vector p1=down[down.size()-2], p2=down.back();
            if (sgn((p2-p1).cross(p-p2)) == 1) break;  // 判断是否符合凸多边形
            down.pop_back();  // 如果不符合，那么最后那个点需要去掉
        }
        down.push_back(p);  // 加入最新的点
    }
    // 上凸包
    vector<Vector> up;
    for (auto it=sets.rbegin();it!=sets.rend();++it) {
        Vector p(it->first, it->second);
        while (up.size()>1) {
            Vector p1=up[up.size()-2], p2=up.back();
            if (sgn((p2-p1).cross(p-p2)) == 1) break;
            up.pop_back();
        }
        up.push_back(p);
    }
    Polygon pol;
    fori(0, down.size()-2) pol.points.push_back(down[i]);  // up down开头和结尾的点是重复的，删掉一个
    fori(0, up.size()-2) pol.points.push_back(up[i]);
    return pol;
}

void solve() {
	int n;
    cin >> n;
    set<pair<double,double>> sets;
    fori(1, n) {
        int x, y;
        cin >> x >> y;
        sets.insert({x, y});
    }
    vector<Vector> other;
    Polygon pol = getConvexHull(sets);

    for (auto pi: sets) {
        Vector vec(pi.first, pi.second);
        bool in = false;
        for (auto v: pol.points) if (vec == v) {
            in = true;
            break;
        }
        if (!in) other.push_back(vec);
    }
    assert(other.size()+pol.points.size() == n);
    int ans = 1;
    fori(0, pol.points.size()-1) {
        Polygon pol2;
        pol2.points.push_back(pol.points[i]);
        pol2.points.push_back(pol.points[(i+1)%pol.points.size()]);
        Line line(pol.points[i], pol.points[(i+1)%pol.points.size()]);
        sort(other.begin(), other.end(), [&](const Vector& v1, const Vector& v2) -> bool {
            return line.distanceToPoint(v1)<line.distanceToPoint(v2);
        });
        forj(0, other.size()-1) {
            pol2.points.push_back(other[j]);
            bool can = true;
            fork(0, j-1) {
                if (pol2.isInPolygon(other[k])) {
                    can = false;
                    break;
                }
            }
            if (can) ans++;
            pol2.points.pop_back();
        }
    }
    cout << ans << endl;
}

signed main() {
	IOS;
	int t = 1;
	while (t--) {
		solve();
	}
	return 0;
}