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;
struct node {
  int x, y;
  long double d;
} a[310], e[310 * 310];
long long f[310][310][6];
int cmp(node x, node y) { return x.d < y.d; }
int main() {
  int n, i, j, l;
  scanf("%d", &n);
  for (i = 1; i <= n; i++) {
    scanf("%d%d", &a[i].x, &a[i].y);
  }
  int tot = 0;
  for (i = 1; i <= n; i++)
    for (j = 1; j <= n; j++)
      if (i != j) {
        e[++tot].x = i, e[tot].y = j;
        e[tot].d = atan2(a[j].y - a[i].y, a[j].x - a[i].x);
      }
  sort(e + 1, e + tot + 1, cmp);
  for (i = 1; i <= n; i++) f[i][i][0] = 1;
  for (i = 1; i <= tot; i++) {
    for (j = 1; j <= n; j++) {
      for (l = 0; l < 5; l++) {
        f[j][e[i].y][l + 1] += f[j][e[i].x][l];
      }
    }
  }
  long long ans = 0;
  for (i = 1; i <= n; i++) ans += f[i][i][5];
  printf("%lld", ans);
  return 0;
}
