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 Point {
  long long x, y;
  Point() {}
  Point(long long x, long long y) : x(x), y(y) {}
  Point operator-(const Point b) const { return Point(x - b.x, y - b.y); }
  long long operator/(const Point b) const { return x * b.y - y * b.x; }
};
Point P[305];
long long n, f[305];
bitset<305> bi[305][305][2];
inline bool same(long long x, long long y) {
  return (x > 0 && y > 0) || (x < 0 && y < 0);
}
inline int count(int i, int j, int k) {
  return (bi[i][j][(P[i] - P[k]) / (P[j] - P[k]) > 0] &
          bi[j][k][(P[j] - P[i]) / (P[k] - P[i]) > 0] &
          bi[i][k][(P[i] - P[j]) / (P[k] - P[j]) > 0])
      .count();
}
int main() {
  ios::sync_with_stdio(false);
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> P[i].x >> P[i].y;
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= n; j++)
      for (int k = 1; k <= n; k++) {
        if (i != j && j != k && i != k) {
          bi[i][j][(P[i] - P[k]) / (P[j] - P[k]) > 0][k] = 1;
        }
      }
  for (int i = 1; i <= n; i++)
    for (int j = i + 1; j <= n; j++)
      for (int k = j + 1; k <= n; k++) f[count(i, j, k)]++;
  long long ans =
      (n * (n - 1) * (n - 2) * (n - 3) * (n - 4)) / 1 / 2 / 3 / 4 / 5 * 2;
  for (int i = 1; i <= n; i++) ans += i * (i - 1) * f[i] - i * f[i] * (n - 4);
  cout << ans / 2 << endl;
  return 0;
}
