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, cnt;
bitset<305> bm[305][305];
long long ans = 1, sub, f[305], x[305], y[305];
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) scanf("%lld%lld", x + i, y + i);
  for (int i = 0; i < n; i++)
    for (int j = 0; j < n; j++)
      if (i != j) {
        long long xa = x[j] - x[i];
        long long ya = y[j] - y[i];
        for (int k = 0; k < n; k++)
          if (i != k && j != k) {
            long long xb = x[k] - x[i];
            long long yb = y[k] - y[i];
            if (xa * yb > xb * ya) bm[i][j].set(k);
          }
      }
  for (int i = 0; i < n; i++)
    for (int j = i + 1; j < n; j++)
      for (int k = j + 1; k < n; k++) {
        if (bm[i][j].test(k))
          cnt = (bm[i][j] & bm[j][k] & bm[k][i]).count();
        else
          cnt = (bm[i][k] & bm[k][j] & bm[j][i]).count();
        f[cnt]++;
      }
  for (int i = 0; i < 5; i++) ans *= (n - i);
  for (int i = 2; i <= 5; i++) ans /= i;
  for (int i = 1; i <= n; i++) {
    sub += i * f[i] * (n - 4);
    ans += i * (i - 1) / 2 * f[i];
  }
  sub /= 2;
  ans -= sub;
  printf("%lld\n", ans);
}
