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 x[310], y[310];
int from[180010], to[180010];
double dir[180010];
long long f[310][310][6];
void qsort(int l, int r) {
  int i = l, j = r, ti;
  double mid = dir[(l + r) / 2], td;
  while (i <= j) {
    while (dir[i] < mid) i++;
    while (dir[j] > mid) j--;
    if (i <= j) {
      ti = from[i];
      from[i] = from[j];
      from[j] = ti;
      ti = to[i];
      to[i] = to[j];
      to[j] = ti;
      td = dir[i];
      dir[i] = dir[j];
      dir[j] = td;
      i++;
      j--;
    }
  }
  if (i < r) qsort(i, r);
  if (l < j) qsort(l, j);
}
int main() {
  memset(f, 0, sizeof(f));
  int n, i, j, k, cnt = 0;
  scanf("%d", &n);
  for (i = 1; i <= n; i++) scanf("%d%d", &x[i], &y[i]);
  for (i = 1; i <= n; i++)
    for (j = 1; j <= n; j++)
      if (i != j) {
        cnt++;
        from[cnt] = i, to[cnt] = j;
        dir[cnt] = atan2(y[j] - y[i], x[j] - x[i]);
        if (dir[cnt] < 0) dir[cnt] += 3.1415926 * 2;
      }
  qsort(1, cnt);
  for (i = 1; i <= n; i++) f[i][i][0] = 1;
  for (i = 1; i <= cnt; i++) {
    for (j = 1; j <= n; j++)
      for (k = 1; k <= 5; k++) f[j][to[i]][k] += f[j][from[i]][k - 1];
  }
  long long ans = 0;
  for (i = 1; i <= n; i++) ans += f[i][i][5];
  printf("%lld", ans);
  return 0;
}
