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;
ifstream fin("test.in");
long long dp[10][305][305];
int x[305], y[305];
pair<double, pair<int, int> > t[90005];
int main() {
  int n, i, j, p, k = 0;
  long long sol = 0;
  cin >> n;
  for (i = 1; i <= n; i++) {
    cin >> x[i] >> y[i];
    for (j = 1; j < i; j++) {
      t[++k] = make_pair(atan2(y[i] - y[j], x[i] - x[j]), make_pair(i, j));
      t[++k] = make_pair(atan2(y[j] - y[i], x[j] - x[i]), make_pair(j, i));
    }
  }
  sort(t + 1, t + k + 1);
  for (i = 1; i <= n; i++) dp[0][i][i] = 1;
  for (i = 1; i <= k; i++) {
    for (j = 1; j <= 5; j++) {
      for (p = 1; p <= n; p++) {
        dp[j][p][t[i].second.second] += dp[j - 1][p][t[i].second.first];
      }
    }
  }
  for (i = 1; i <= n; i++) sol = sol + dp[5][i][i];
  cout << sol;
}
