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 = 3e2 + 10;
struct point {
  int x, y;
  point operator-(const point &rhs) const {
    return (point){x - rhs.x, y - rhs.y};
  }
  long long operator*(const point &rhs) const {
    return 1ll * y * rhs.x - 1ll * x * rhs.y;
  }
} p[maxn];
int n;
bitset<maxn> halfp[maxn][maxn];
int main(void) {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%d%d", &p[i].x, &p[i].y);
  }
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) {
      for (int k = 0; k < n; k++) {
        if (i != j && i != k && k != j && (p[i] - p[k]) * (p[j] - p[k]) > 0) {
          halfp[i][j].set(k);
        }
      }
    }
  }
  long long ans = 2;
  for (int i = 1; i <= 5; i++) {
    ans = ans * (n - i + 1) / i;
  }
  for (int i = 0; i < n; i++) {
    for (int j = i + 1; j < n; j++) {
      for (int k = i + 1; k < n; k++) {
        if (j != k) {
          int cur = (halfp[i][j] & halfp[j][k] & halfp[k][i]).count();
          ans -= 1ll * cur * (n - 3 - cur);
        }
      }
    }
  }
  cout << ans / 2 << endl;
  return 0;
}
