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;
long long n, m, ans, x[320], y[320], f[6][320][320];
struct node {
  long long x, y, u, v;
  node(long long x = 0, long long y = 0, long long u = 0, long long v = 0)
      : x(x), y(y), u(u), v(v) {}
  bool operator<(const node &w) const {
    long long o1 = x > 0 || (x == 0 && y > 0),
              o2 = (w.x > 0) || (w.x == 0 && w.y > 0);
    if (o1 != o2) return o1 < o2;
    return x * w.y - y * w.x >= 0;
  }
} a[320 * 320];
int main() {
  scanf("%lld", &n);
  for (long long i = 1; i <= n; ++i) {
    scanf("%lld%lld", x + i, y + i);
  }
  for (long long i = 1; i <= n; ++i)
    for (long long j = 1; j <= n; ++j)
      if (i != j) a[++m] = node(x[i] - x[j], y[i] - y[j], i, j);
  sort(a + 1, a + m + 1);
  for (long long i = 1; i <= n; ++i) f[0][i][i] = 1;
  for (long long i = 1; i <= m; ++i)
    for (long long j = 0; j < 5; ++j)
      for (long long k = 1; k <= n; ++k) f[j + 1][k][a[i].v] += f[j][k][a[i].u];
  for (long long i = 1; i <= n; ++i) ans += f[5][i][i];
  printf("%lld\n", ans);
  return 0;
}
