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>
#pragma GCC optimization("O3")
#pragma GCC optimization("unroll-loops")
using namespace std;
pair<long long, long long> operator-(const pair<long long, long long> &a,
                                     const pair<long long, long long> &b) {
  return {a.first - b.first, a.second - b.second};
}
long long cross(const pair<long long, long long> &a,
                const pair<long long, long long> &b) {
  return a.first * b.second - a.second * b.first;
}
long long cross(const pair<long long, long long> &a, const pair<int, int> &b,
                const pair<long long, long long> &c) {
  return cross(a - c, b - c);
}
bitset<305> ok[305][305];
int main() {
  int n;
  scanf("%d", &n);
  pair<long long, long long> p[305];
  for (int i = 0; i < n; i++) {
    int first, second;
    scanf("%d %d", &first, &second);
    p[i] = make_pair((first), (second));
  }
  sort(p, p + n);
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) {
      if (i != j) {
        for (int k = 0; k < n; k++) {
          if (k != i && k != j) {
            if (cross(p[j], p[k], p[i]) > 0) {
              ok[i][j][k] = 1;
            }
          }
        }
      }
    }
  }
  long long ans = 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 (j != k && cross(p[j], p[k], p[i]) > 0) {
          int a = (ok[i][j] & ok[i][k] & ok[j][k]).count();
          int b = (ok[i][j] & ok[k][j] & ok[k][i]).count();
          int c = (ok[j][i] & ok[j][k] & ok[k][i]).count();
          ans += c * a;
        }
      }
    }
    for (int j = 0; j < n; j++)
      for (int k = 0; k < n; k++) ok[j][k][i] = 0;
  }
  printf("%lld\n", ans);
}
