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 = 305;
int n, ct, x[Maxn], y[Maxn], ord[Maxn];
long long ans, f[5][Maxn][Maxn];
long long cross(int a, int b, int c) {
  return (x[a] - x[b]) * (long long)(y[b] - y[c]) -
         (x[b] - x[c]) * (long long)(y[a] - y[b]);
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d%d", &x[i], &y[i]);
  for (int i = 1; i <= n; i++) {
    ct = 0;
    for (int c = 2; c <= 4; c++)
      for (int j = 1; j <= n; j++)
        for (int k = 1; k < j; k++) f[c][j][k] = 0;
    for (int j = 1; j <= n; j++)
      if (make_pair(y[j], x[j]) > make_pair(y[i], x[i])) ord[++ct] = j;
    sort(ord + 1, ord + 1 + ct, [&](int a, int b) {
      return atan2(y[a] - y[i], x[a] - x[i]) < atan2(y[b] - y[i], x[b] - x[i]);
    });
    for (int j = 1; j <= ct; j++) f[1][j][0] = 1;
    for (int c = 2; c <= 4; c++)
      for (int j = 1; j <= ct; j++)
        for (int k = (c != 2); k < j; k++)
          if (f[c - 1][j][k])
            for (int q = j + 1; q <= ct; q++)
              if (cross(c == 2 ? i : ord[k], ord[j], ord[q]) > 0)
                f[c][q][j] += f[c - 1][j][k];
    for (int j = 1; j <= ct; j++)
      for (int k = 1; k < j; k++)
        if (cross(ord[k], ord[j], i) > 0) ans += f[4][j][k];
  }
  printf("%lld", ans);
  return 0;
}
