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 N = 310;
int n, tot;
long long ans, f[N][N][5];
struct EDGE {
  int x, y;
  double d;
} g[N], e[N * N];
int cmp(EDGE a, EDGE b) { return a.d < b.d; }
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d %d", &g[i].x, &g[i].y);
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= n; j++)
      if (i != j)
        e[++tot] = (EDGE){i, j, atan2(g[j].y - g[i].y, g[j].x - g[i].x)};
  sort(e + 1, e + 1 + tot, cmp);
  for (int i = 1; i <= tot; i++) {
    f[e[i].x][e[i].y][0] = 1;
    for (int j = 1; j <= n; j++)
      for (int k = 1; k <= 4; k++) f[j][e[i].y][k] += f[j][e[i].x][k - 1];
  }
  for (int i = 1; i <= n; i++) ans += f[i][i][4];
  printf("%lld\n", ans);
  return 0;
}
