#include<bits/stdc++.h>
using namespace std;

#define int __int128

void read(int& n) {
	int x = 0, f = 1;
	char ch = getchar();
	while (ch < '0' || ch>'9') {
		if (ch == '-') f = -1;
		ch = getchar();
	}
	while (ch >= '0' && ch <= '9') {
		x = (x << 1) + (x << 3) + (ch ^ 48);
		ch = getchar();
	}
	n = x * f;
}
void print(int n) {
	if (n < 0) {
		putchar('-');
		n *= -1;
	}
	if (n > 9) print(n / 10);
	putchar(n % 10 + '0');
}
//#undef int

struct Point
{
	int x;
	int y;
	int f;
}d1[3005],d2[3005];
int vs[3005],st[3005],cnt,n;

int cha(Point a, Point b, Point c) {
	int xa = b.x - a.x;
	int ya = b.y - a.y;
	int xb = c.x - a.x;
	int yb = c.y - a.y;
	return yb * xa - ya * xb;
}

bool cmp1(Point a, Point b) {
	if (a.x == b.x) return a.y < b.y;
	else return a.x < b.x;
}

void TUBAO() {
	st[1] = 1; st[2] = 2; cnt = 2;
	for (int i = 3; i <= n; i++) {
		while (cnt > 1 && cha(d1[st[cnt - 1]], d1[st[cnt]], d1[i]) <= 0) cnt--;
		st[++cnt] = i;
	}
	st[++cnt] = n - 1;
	for (int i = n - 2; i >= 1; i--) {
		while (cnt > 1 && cha(d1[st[cnt - 1]], d1[st[cnt]], d1[i]) <= 0) cnt--;
		st[++cnt] = i;
	}
	--cnt;//最后会重复
}

int Cross(Point a, Point ots)
{
	return a.x * ots.y - a.y * ots.x;
}

int Quadrantt(Point a)//象限排序，注意包含四个坐标轴
{
	if (a.x > 0 && a.y >= 0)  return 1;
	if (a.x <= 0 && a.y > 0)  return 2;
	if (a.x < 0 && a.y <= 0)  return 3;
	if (a.x >= 0 && a.y < 0)  return 4;
	return 0;
}

bool cmp2(Point a, Point b)  //先按象限从小到大排序 再按极角从小到大排序
{
	if (Quadrantt(a) == Quadrantt(b))//返回值就是象限
		return Cross(a, b) < 0;//同一个象限计算叉乘   
	else return Quadrantt(a) < Quadrantt(b);
}

signed main()
{
	//ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
	read(n);
	for (int i = 1; i <= n; i++) read(d1[i].x),read(d1[i].y);
	sort(d1 + 1, d1 + 1 + n, cmp1);
//	for(int i=1;i<=n;i++){
//		print(d1[i].x);
//		cout<<" ";
//		print(d1[i].y);
//		cout<<"\n";
//	} 
	TUBAO();
	int ans = 1;//凸包算一个
	
	for (int i = 1; i <=cnt; i++) vs[st[i]] = 1;
	int tot = 0;
	for (int i = 1; i <= n; i++) {
		if (vs[i]) continue;//只能选非凸包的点
		tot = 0;
		for (int j = 1; j <= n; j++) {
			if (j == i) continue;
			d2[++tot] = { d1[j].x - d1[i].x,d1[j].y - d1[i].y,vs[j]};
			//if (vs[j]) d2[tot].f = 1;//打标记
		}
		sort(d2 + 1, d2 + 1 + n - 1, cmp2);
		for (int j = 1; j < n - 1; j++) {
			if (d2[j].f && d2[j + 1].f) ans++;
		}
		if (d2[n - 1].f && d2[1].f) ans++;
	}
	print(ans);
	return 0;
}