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;
int n, f[301][301][301], cnt[2], s[301][301], nw;
struct P {
  int x, y;
} a[311];
long long ans, sum;
bool cmp(P a, P b) { return a.x < b.x; }
P operator-(P a, P b) { return (P){a.x - b.x, a.y - b.y}; }
long long operator^(P a, P b) { return 1ll * a.x * b.y - 1ll * a.y * b.x; }
int main() {
  scanf("%d", &n), ans = 1ll * n * (n - 1) * (n - 2) * (n - 3) * (n - 4) / 120;
  for (int i = 1; i <= n; i++) scanf("%d%d", &a[i].x, &a[i].y);
  sort(a + 1, a + 1 + n, cmp);
  for (int i = 1; i <= n; i++)
    for (int j = i + 1; j <= n; j++) {
      cnt[0] = cnt[1] = 0;
      for (int k = 1; k <= n; k++)
        if (i != k && j != k) cnt[((a[j] - a[i]) ^ (a[k] - a[i])) > 0]++;
      for (int k = 1; k < i; k++)
        f[k][i][j] += cnt[((a[j] - a[i]) ^ (a[k] - a[i])) < 0];
      for (int k = i + 1; k < j; k++)
        f[i][k][j] += cnt[((a[j] - a[i]) ^ (a[k] - a[i])) < 0];
      for (int k = j + 1; k <= n; k++)
        f[i][j][k] += cnt[((a[j] - a[i]) ^ (a[k] - a[i])) < 0];
      for (int k = 1; k <= n; k++)
        if (i != k && j != k && a[k].x >= a[i].x && a[k].x <= a[j].x)
          s[i][j] += (((a[j] - a[i]) ^ (a[k] - a[i])) < 0);
    }
  for (int i = 1; i <= n; i++)
    for (int j = i + 1; j <= n; j++)
      for (int k = j + 1; k <= n; k++) {
        if (a[i].x == a[j].x)
          nw = abs(s[i][k] - s[j][k]) - 1;
        else if (a[j].x == a[k].x)
          nw = abs(s[i][j] - s[i][k]) - 1;
        else {
          nw = s[i][j] + s[j][k] - s[i][k];
          if (nw < 0) nw = -nw - 1;
        }
        sum += 1ll * f[i][j][k] * nw;
        ans += 1ll * nw * (nw - 1) / 2;
      }
  return printf("%I64d\n", ans - sum / 2), 0;
}
