You are given a set of n points in a 2D plane. No three points are collinear.

A pentagram is a set of 5 points A,B,C,D,E that can be arranged as follows. <image> Note the length of the line segments don't matter, only that those particular intersections exist.

Count the number of ways to choose 5 points from the given set that form a pentagram.

Input

The first line contains an integer n (5 ≤ n ≤ 300) — the number of points.

Each of the next n lines contains two integers x_i, y_i (-10^6 ≤ x_i,y_i ≤ 10^6) — the coordinates of the i-th point. It is guaranteed that no three points are collinear.

Output

Print a single integer, the number of sets of 5 points that form a pentagram.

Examples

Input


5
0 0
0 2
2 0
2 2
1 3


Output


1


Input


5
0 0
4 0
0 4
4 4
2 3


Output


0


Input


10
841746 527518
595261 331297
-946901 129987
670374 -140388
-684770 309555
-302589 415564
-387435 613331
-624940 -95922
945847 -199224
24636 -565799


Output


85

Note

A picture of the first sample: <image> A picture of the second sample: <image> A picture of the third sample: <image>
#include <bits/stdc++.h>
using namespace std;
const int maxn = 320;
struct point {
  long long x, y;
  point() {}
  point(long long a, long long b) : x(a), y(b){};
  point operator-(point a) { return point(x - a.x, y - a.y); }
  bool operator<(point a) const {
    if (x == a.x) return y < a.y;
    return x < a.x;
  }
} dt[maxn];
struct line {
  double ang;
  int id1, id2;
  bool operator<(line a) const { return ang < a.ang; }
} V[maxn * maxn];
int n, tot;
long long f[maxn][maxn][6];
long long xmul(point a, point b) { return a.x * b.y - a.y * b.x; }
void init() {
  for (int i = 1; i <= n; i++) {
    for (int j = i + 1; j <= n; j++) {
      V[++tot] = (line){atan2(dt[j].y - dt[i].y, dt[j].x - dt[i].x), i, j};
      V[++tot] = (line){atan2(dt[i].y - dt[j].y, dt[i].x - dt[j].x), j, i};
    }
  }
  sort(V + 1, V + tot + 1);
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%lld %lld", &dt[i].x, &dt[i].y);
  init();
  for (int i = 1; i <= n; i++) f[i][i][0] = 1;
  for (int i = 1; i <= tot; i++) {
    int x = V[i].id1, y = V[i].id2;
    for (int j = 1; j <= n; j++) {
      for (int k = 0; k <= 4; k++) {
        f[j][y][k + 1] += f[j][x][k];
      }
    }
  }
  long long ans = 0;
  for (int i = 1; i <= n; i++) ans += f[i][i][5];
  cout << ans << endl;
}
