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;
using point = complex<long long>;
long long cross(point a, point b) {
  return a.real() * b.imag() - a.imag() * b.real();
}
const int N = 300;
int under[N][N];
point points[N];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  int n;
  cin >> n;
  for (int i = 0; i < n; ++i) {
    int x, y;
    cin >> x >> y;
    points[i] = point(x, y);
  }
  sort(points, points + n, [&](point a, point b) {
    if (a.real() == b.real()) {
      return a.imag() < b.imag();
    }
    return a.real() < b.real();
  });
  for (int i = 0; i < n; ++i) {
    for (int j = i + 1; j < n; ++j) {
      for (int k = j + 1; k < n; ++k) {
        under[i][k] +=
            (cross(points[j] - points[i], points[k] - points[i]) > 0);
      }
    }
  }
  long long ans = 2;
  for (int i = 0; i < 5; ++i) {
    ans *= n - i;
    ans /= i + 1;
  }
  for (int i = 0; i < n; ++i) {
    for (int j = i + 1; j < n; ++j) {
      for (int k = j + 1; k < n; ++k) {
        int inside =
            abs(under[i][k] - under[i][j] - under[j][k] -
                (cross(points[j] - points[i], points[k] - points[i]) > 0));
        ans -= inside * (n - inside - 3);
      }
    }
  }
  cout << ans / 2 << "\n";
}
