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;
struct point {
  int x, y;
  point(int x = 0, int y = 0) : x(x), y(y) {}
  point operator-(const point &p) const { return point(x - p.x, y - p.y); }
  long long operator*(const point &p) const {
    return (long long)x * p.y - (long long)y * p.x;
  }
};
long long ccw(point o, point u, point v) { return (u - o) * (v - o); }
const int N = 305;
point a[N];
bitset<N> f[N][N];
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) {
    scanf("%d %d", &a[i].x, &a[i].y);
  }
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < n; ++j) {
      for (int k = 0; k < n; ++k) {
        f[i][j][k] = (ccw(a[i], a[j], a[k]) > 0);
      }
    }
  }
  long long ans = (long long)n * (n - 1) * (n - 2) * (n - 3) * (n - 4) / 120;
  long long sum = 0;
  for (int i = 0; i < n; ++i) {
    for (int j = i + 1; j < n; ++j) {
      for (int k = i + 1; k < n; ++k) {
        if (f[i][j][k]) {
          int cur = (f[i][j] & f[j][k] & f[k][i]).count();
          sum += cur * (n - 4);
          ans += cur * (cur - 1) / 2;
        }
      }
    }
  }
  printf("%lld\n", ans - sum / 2);
}
