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 N = 300;
struct point {
  int x, y;
  point(int x = 0, int y = 0) : x(x), y(y) {}
  point operator-(const point &other) const {
    return point(x - other.x, y - other.y);
  }
  long long operator*(const point &other) const {
    return (long long)x * other.y - (long long)y * other.x;
  }
};
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  vector<point> p(n);
  for (int i = 0; i < n; ++i) {
    cin >> p[i].x >> p[i].y;
  }
  vector<vector<bitset<N>>> left(n, vector<bitset<N>>(n));
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < n; ++j) {
      if (j != i) {
        for (int k = 0; k < n; ++k) {
          if (k != i && k != j) {
            left[i][j][k] = (p[j] - p[i]) * (p[k] - p[i]) > 0;
          }
        }
      }
    }
  }
  long long foo = 0, bar = 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 (left[i][j][k]) {
          int inside = (left[i][j] & left[j][k] & left[k][i]).count();
          foo += inside;
          bar += inside * (inside - 1) / 2;
        }
      }
    }
  }
  cout << (long long)n * (n - 1) * (n - 2) * (n - 3) * (n - 4) / 120 -
              foo * (n - 4) / 2 + bar
       << "\n";
  return 0;
}
