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;
const int N = 310;
int gi() {
  int x = 0, o = 1;
  char ch = getchar();
  while ((ch < '0' || ch > '9') && ch != '-') ch = getchar();
  if (ch == '-') o = -1, ch = getchar();
  while (ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar();
  return x * o;
}
int n, x[N], y[N];
long long f[N][N][5], ans = 0;
struct data {
  int a, b;
  double c;
  bool operator<(const data &A) const { return c < A.c; }
};
data st[N * N];
int tp = 0;
int main() {
  n = gi();
  for (int i = 1; i <= n; i++) x[i] = gi(), y[i] = gi();
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= n; j++)
      if (i ^ j) st[++tp] = (data){i, j, atan2(y[j] - y[i], x[j] - x[i])};
  sort(st + 1, st + tp + 1);
  for (int i = 1; i <= tp; i++) {
    int a = st[i].a, b = st[i].b;
    f[a][b][0] = 1;
    for (int s = 1; s <= n; s++)
      for (int j = 0; j < 4; j++) f[s][b][j + 1] += f[s][a][j];
  }
  for (int i = 1; i <= n; i++) ans += f[i][i][4];
  printf("%lld\n", ans);
  return 0;
}
