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 = 300 + 5;
long long dp[maxn][maxn][6];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n;
  cin >> n;
  vector<complex<double>> P(n);
  for (int i = 0, x, y; i < n; i++) {
    cin >> x >> y;
    P[i] = complex<double>(1.0 * x, 1.0 * y);
  }
  vector<pair<double, pair<int, int>>> p;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) {
      if (i != j) {
        complex<double> v = P[j] - P[i];
        p.push_back({arg(v), {i, j}});
      }
    }
  }
  for (int i = 0; i < n; i++) dp[i][i][0] = 1;
  sort(p.begin(), p.end());
  for (int i = 0; i < (int)p.size(); i++) {
    int a = p[i].second.first;
    int b = p[i].second.second;
    for (int j = 0; j < n; j++)
      for (int k = 0; k < 5; k++) dp[j][b][k + 1] += dp[j][a][k];
  }
  long long ans = 0;
  for (int i = 0; i < n; i++) ans += dp[i][i][5];
  cout << ans << '\n';
  return 0;
}
