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;
long long n, xx, ans, dp[310][310][6];
double x[310], y[310];
struct Xd {
  long long u, v;
  double jd;
  bool operator<(const Xd &u) const { return jd < u.jd; }
} xd[310 * 310];
int main() {
  long long i, j, k;
  cin >> n;
  for (i = 1; i <= n; i++) scanf("%lf%lf", &x[i], &y[i]);
  for (i = 1; i <= n; i++) {
    for (j = 1; j <= n; j++) {
      if (i == j) continue;
      xd[++xx] = (Xd){i, j, atan2(y[j] - y[i], x[j] - x[i])};
    }
  }
  sort(xd + 1, xd + xx + 1);
  for (i = 1; i <= xx; i++) {
    dp[xd[i].u][xd[i].v][1] = 1;
    for (j = 1; j <= n; j++) {
      for (k = 2; k <= 5; k++) {
        dp[j][xd[i].v][k] += dp[j][xd[i].u][k - 1];
      }
    }
  }
  for (i = 1; i <= n; i++) ans += dp[i][i][5];
  cout << ans;
}
