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, MAXM = 100005;
bitset<MAXN> bt[MAXM];
int id[MAXN][MAXN], n;
long long xx[MAXN], yy[MAXN], f[MAXN];
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%lld%lld", xx + i, yy + i);
  int cnt = 0;
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= n; j++)
      if (i != j) {
        id[i][j] = ++cnt;
        for (int k = 1; k <= n; k++)
          if (k != i && k != j)
            if ((xx[j] - xx[i]) * (yy[k] - yy[i]) -
                    (xx[k] - xx[i]) * (yy[j] - yy[i]) >
                0)
              bt[cnt].set(k);
      }
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= n; j++)
      if (i != j)
        for (int k = 1; k <= n; k++)
          if (i != k && j != k)
            if ((xx[j] - xx[i]) * (yy[k] - yy[i]) -
                    (xx[k] - xx[i]) * (yy[j] - yy[i]) >
                0)
              ++f[(bt[id[i][j]] & bt[id[j][k]] & bt[id[k][i]]).count()];
  long long res = 0;
  for (int i = 0; i <= n; i++) f[i] /= 3;
  for (int i = 1; i <= n; i++) res += i * f[i] * (n - 4);
  for (int i = 1; i <= n; i++) res -= i * (i - 1) * f[i];
  printf("%lld\n",
         (long long)n * (n - 1) * (n - 2) * (n - 3) * (n - 4) / 120 - res / 2);
  return 0;
}
