#include <bits/stdc++.h>
using namespace std;
typedef pair<int, int> PII;
const int N = 2010;
struct Point { 
	double x, y; 
	int id;
	bool operator< (const Point& r){
		if(x != r.x) return x < r.x;
		else return y < r.y;
	}
};        // 点
using Vec = Point;                    // 向量
struct Line { Point P; Vec v; };      // 直线（点向式）
struct Seg { Point A, B; };           // 线段（存两个端点）
struct Circle { Point O; double r; }; // 圆（存圆心和半径）
Point a[N];
using Points = vector<Point>;
const Point O = {0, 0};                        // 原点
const Line Ox = {O, {1, 0}}, Oy = {O, {0, 1}}; // 坐标轴
const double PI = acos(-1), EPS = 1e-9;
bool eq(double a, double b) { return abs(a - b) < EPS; } // ==
bool gt(double a, double b) { return a - b > EPS; }      // >
bool lt(double a, double b) { return a - b < -EPS; }     // <
bool ge(double a, double b) { return a - b > -EPS; }     // >=
bool le(double a, double b) { return a - b < EPS; }      // <=
Vec operator+(Vec u, Vec v) { return {u.x + v.x, u.y + v.y}; }   // 向量加向量
Vec operator-(Vec u, Vec v) { return {u.x - v.x, u.y - v.y}; }   // 向量减向量
Vec operator*(double k, Vec v) { return {k * v.x, k * v.y}; }    // 数乘
double operator*(Vec u, Vec v) { return u.x * v.x + u.y * v.y; } // 点乘
double cross(Vec u, Vec v) { return u.x * v.y - u.y * v.x; } // 叉乘
bool check(Point p, Point q, Point r) // 检查三个点组成的两个向量的旋转方向是否为逆时针
{
    return lt(0, cross(q - p, r - q));
}
Points chull(Points &ps)
{
    sort(ps.begin(), ps.end());
    vector<int> I{0}, used(ps.size());
    for (int i = 1; i < ps.size(); i++)
    {
        while (I.size() > 1 && !check(ps[I[I.size() - 2]], ps[I.back()], ps[i]))
            used[I.back()] = 0, I.pop_back();
        used[i] = 1, I.push_back(i);
    }
    for (int i = ps.size() - 2; i >= 0; i--)
    {
        if (used[i])
            continue;
        while (I.size() > 1 && !check(ps[I[I.size() - 2]], ps[I.back()], ps[i]))
            I.pop_back();
        I.push_back(i);
    }
    Points H;
    for (int i = 0; i < I.size() - 1; i++)
        H.push_back(ps[I[i]]);
    return H;
}

double the[2010][2010];
double theta(auto p) { return atan2(p.y, p.x); } // 求极角
void psort(Points &ps, Point c = O)              // 极角排序
{
    sort(ps.begin(), ps.end(), [&](auto p1, auto p2) {
        return lt(the[p1.id][c.id], the[p2.id][c.id]);
    });
}
int n, st[N];
Points v, v1;
int main(){
	cin >> n;
	for(int i=1;i<=n;i++){
		cin >> a[i].x >> a[i].y;
		a[i].id = i;
		v1.push_back(a[i]);
	}
    for(int i = 1; i <= n; i++) {
        for(int j = 1; j <= n; j++) {
            the[i][j] = theta(a[i] - a[j]);
        }
    }
	v = chull(v1);
	for(int i = 0; i < v.size(); i++) st[v[i].id] = 1;
	//for(int i=0;i<v.size;i++) cout << v[i].id;
	int ans = 0;
	for(int i=1;i<=n;i++){
		if(!st[i]){
			Points tmp;
			for(int j=1;j<=n;j++){
				if(j == i) continue;
				tmp.push_back(a[j]);
			}
			psort(tmp, a[i]);
			for(int j=0;j<tmp.size()-1;j++){
				if(st[tmp[j].id] && st[tmp[j+1].id]) ans++;
			}
			if(st[tmp[0].id] && st[tmp[tmp.size()-1].id]) ans++;
		}
	}
	cout << ans + 1 << endl;
	return 0;
}