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>
inline long long gi() {
  long long x = 0, f = 1;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (isdigit(ch)) x = x * 10 + ch - '0', ch = getchar();
  return x * f;
}
int x[310], y[310];
struct edge {
  int x, y;
  double d;
} s[100010];
inline bool operator<(const edge& a, const edge& b) { return a.d < b.d; }
int f[310][310][5];
int main() {
  int n = gi();
  for (int i = 1; i <= n; ++i) x[i] = gi(), y[i] = gi();
  int m = 0;
  for (int i = 1; i <= n; ++i)
    for (int j = 1; j <= n; ++j)
      if (i != j) s[++m] = (edge){i, j, atan2(y[j] - y[i], x[j] - x[i])};
  std::sort(s + 1, s + m + 1);
  long long ans = 0;
  for (int i = 1; i <= m; ++i) {
    int x = s[i].x, y = s[i].y;
    f[x][y][0] = 1;
    for (int a = 1; a <= n; ++a)
      for (int b = 0; b < 4; ++b) f[a][y][b + 1] += f[a][x][b];
  }
  for (int i = 1; i <= n; ++i) ans += f[i][i][4];
  printf("%lld\n", ans);
  return 0;
}
