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;
void read(int &x) {
  char ch = getchar();
  x = 0;
  int bz = 1;
  while (ch < '0' || ch > '9') {
    if (ch == '-') bz = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9')
    x = (x << 1) + (x << 3) + ch - 48, ch = getchar();
  x *= bz;
}
struct EDGE {
  int u, v, x1, y1, x2, y2;
  double k;
} p[90000 << 1];
struct Point {
  int x, y;
} a[300 + 1];
long long f[300 + 1][300 + 1][6];
int n, cnt = 0;
void Add(int u, int v, int x1, int y1, int x2, int y2) {
  p[++cnt] = (EDGE){u, v, x1, y1, x2, y2, atan2(y2 - y1, x2 - x1)};
}
void Link(int u, int v, int x1, int y1, int x2, int y2) {
  Add(u, v, x1, y1, x2, y2), Add(v, u, x2, y2, x1, y1);
}
bool Cmp(EDGE a, EDGE b) { return a.k < b.k; }
int main() {
  read(n);
  for (int i = 1; i <= n; i++) read(a[i].x), read(a[i].y);
  for (int i = 1; i <= n; i++)
    for (int j = i + 1; j <= n; j++) Link(i, j, a[i].x, a[i].y, a[j].x, a[j].y);
  sort(p + 1, p + 1 + cnt, Cmp);
  for (int i = 1; i <= n; i++) f[i][i][0] = 1;
  for (int i = 1; i <= cnt; i++)
    for (int j = 1; j <= n; j++)
      for (int k = 0; k <= 4; k++) f[j][p[i].v][k + 1] += f[j][p[i].u][k];
  long long ans = 0;
  for (int i = 1; i <= n; i++) ans += f[i][i][5];
  printf("%lld\n", ans);
  return 0;
}
