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>
int N;
struct Ad {
  int x, y;
  inline Ad operator-(const Ad &c) const { return (Ad){x - c.x, y - c.y}; }
  inline long long operator*(const Ad &c) const {
    return (long long)x * c.y - (long long)y * c.x;
  }
} A[305];
int C[305][305];
long long Ans;
int main() {
  scanf("%d", &N);
  for (int i = 0; i < N; ++i) scanf("%d%d", &A[i].x, &A[i].y);
  std::sort(A, A + N,
            [](Ad i, Ad j) { return i.x ^ j.x ? i.x < j.x : i.y < j.y; });
  for (int i = 0; i < N; ++i)
    for (int j = i + 1; j < N; ++j)
      for (int k = j + 1; k < N; ++k)
        if ((A[j] - A[i]) * (A[k] - A[i]) > 0) ++C[i][k];
  for (int i = 0; i < N; ++i)
    for (int j = i + 1; j < N; ++j)
      for (int k = j + 1, x; k < N; ++k)
        x = std::abs(C[i][k] - C[i][j] - C[j][k] -
                     ((A[j] - A[i]) * (A[k] - A[i]) > 0)),
        Ans += x * (N - 3 - x);
  return printf("%I64d", (long long)N * (N - 1) / 2 * (N - 2) / 3 * (N - 3) /
                                 4 * (N - 4) / 5 -
                             Ans / 2),
         0;
}
