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 pt {
  long long x, y;
  bool operator<(const pt& b) const {
    if (x != b.x) return x < b.x;
    return y < b.y;
  }
};
long long det(pt a, pt b, pt c) {
  return a.x * b.y + b.x * c.y + c.x * a.y - a.y * b.x - b.y * c.x - c.y * a.x;
}
struct radial {
  pt b;
  bool operator()(pt a, pt c) { return det(b, a, c) < 0; }
};
int n;
pt a[305], b[305];
long long dp[4][305][305];
long long z;
long long daj_dp(pt baza, int n) {
  b[0] = baza;
  sort(b + 1, b + n + 1, radial{baza});
  memset(dp, 0, sizeof(dp));
  for (int i = 1; i <= n; i++) {
    dp[0][0][i] = 1;
  }
  for (int t = 0; t < 3; t++) {
    for (int i = 0; i <= n; i++) {
      for (int j = i + 1; j <= n; j++) {
        if (!dp[t][i][j]) continue;
        for (int k = j + 1; k <= n; k++) {
          if (det(b[i], b[j], b[k]) < 0) dp[t + 1][j][k] += dp[t][i][j];
        }
      }
    }
  }
  long long z = 0;
  for (int i = 0; i <= n; i++) {
    for (int j = i + 1; j <= n; j++) {
      z += dp[3][i][j];
    }
  }
  return z;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  cerr.tie(nullptr);
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> a[i].x >> a[i].y;
  }
  sort(a, a + n);
  long long z = 0;
  for (int i = 0; i < n; i++) {
    copy(a + i + 1, a + n, b + 1);
    z += daj_dp(a[i], n - i - 1);
  }
  cout << z << '\n';
}
