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 {
  long long x, y;
  double ang;
};
struct vec {
  double ang;
  int i, j;
  vec(long long x, long long y, int _i, int _j) : i(_i), j(_j) {
    ang = atan2(y, x);
  }
};
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int n;
  cin >> n;
  vector<point> a;
  for (int i = 0; i < n; i++) {
    long long x, y;
    cin >> x >> y;
    a.push_back({x, y});
  }
  vector<vec> v;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) {
      if (i != j) {
        v.push_back(vec(a[j].x - a[i].x, a[j].y - a[i].y, i, j));
      }
    }
  }
  sort(v.begin(), v.end(),
       [](const vec &a, const vec &b) { return a.ang < b.ang; });
  long long dp[n][n][5];
  memset(dp, 0, sizeof dp);
  for (auto &vec : v) {
    dp[vec.i][vec.j][0] = 1;
    for (int i = 0; i < 4; i++) {
      for (int j = 0; j < n; j++) {
        dp[j][vec.j][i + 1] += dp[j][vec.i][i];
      }
    }
  }
  long long ans = 0;
  for (int i = 0; i < n; i++) {
    ans += dp[i][i][4];
  }
  cout << ans << '\n';
  return 0;
}
