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 = 310;
int n, x[maxn], y[maxn], cnt[maxn][maxn][maxn];
bitset<maxn> tmp, b[maxn][maxn];
long long ans1, ans2;
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d %d", &x[i], &y[i]);
  }
  ans1 = 1LL * n * (n - 1) * (n - 2) * (n - 3) * (n - 4) / 120;
  auto cross = [&](int i, int j, int k) {
    int x1 = x[j] - x[i], y1 = y[j] - y[i];
    int x2 = x[k] - x[i], y2 = y[k] - y[i];
    return 1LL * x1 * y2 - 1LL * x2 * y1;
  };
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++) {
      for (int k = 1; k <= n; k++)
        if (i ^ k && j ^ k) {
          if (cross(i, j, k) < 0) b[i][j][k] = 1;
        }
    }
  }
  for (int i = 1; i <= n; i++) {
    for (int j = i + 1; j <= n; j++) {
      for (int k = j + 1; k <= n; k++) {
        tmp = cross(i, j, k) < 0 ? b[i][j] : b[j][i];
        tmp &= cross(j, k, i) < 0 ? b[j][k] : b[k][j];
        tmp &= cross(i, k, j) < 0 ? b[i][k] : b[k][i];
        cnt[i][j][k] = tmp.count();
        ans1 -= 1LL * cnt[i][j][k] * (cnt[i][j][k] - 1) / 2;
      }
    }
  }
  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) {
            vector<int> V = {i, j, k};
            sort(V.begin(), V.end());
            cnt[i][j][k] = cnt[V[0]][V[1]][V[2]];
          }
      }
  }
  for (int i = 1; i <= n; i++) {
    for (int j = i + 1; j <= n; j++) {
      bitset<maxn> b1, b2;
      for (int k = 1; k <= n; k++)
        if (i ^ k && j ^ k) {
          if (cross(i, j, k) < 0)
            b1.set(k);
          else
            b2.set(k);
        }
      for (int k = 1; k <= n; k++)
        if (i ^ k && j ^ k) {
          if (cross(i, j, k) < 0) {
            tmp = b2;
            tmp &= cross(i, k, j) < 0 ? b[i][k] : b[k][i];
            tmp &= cross(j, k, i) < 0 ? b[j][k] : b[k][j];
            ans2 += 1LL * tmp.count() * cnt[i][j][k];
          } else {
            tmp = b1;
            tmp &= cross(i, k, j) < 0 ? b[i][k] : b[k][i];
            tmp &= cross(j, k, i) < 0 ? b[j][k] : b[k][j];
            ans2 += 1LL * tmp.count() * cnt[i][j][k];
          }
        }
    }
  }
  printf("%lld\n", ans1 - ans2 / 2);
  return 0;
}
