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 double pi = acos(-1);
struct node {
  int x, y;
  double angle;
} e[310 * 310];
int n, x[310], y[310], tot = 0;
long long ans = 0, f[310][310][6];
inline int read() {
  int x = 0, f = 0;
  char c = getchar();
  while (c < '0' || c > '9') f = (c == '-') ? 1 : f, c = getchar();
  while (c >= '0' && c <= '9')
    x = (x << 1) + (x << 3) + (c ^ 48), c = getchar();
  return f ? -x : x;
}
inline bool cmp(node x, node y) { return x.angle < y.angle; }
int main() {
  n = read();
  for (int i = (1); i <= (n); i++) x[i] = read(), y[i] = read();
  for (int i = (1); i <= (n); i++)
    for (int j = (1); j <= (n); j++) {
      if (i == j) continue;
      e[++tot].x = i, e[tot].y = j;
      e[tot].angle = atan2(y[j] - y[i], x[j] - x[i]);
      if (e[tot].angle < 0) e[tot].angle += 2 * pi;
    }
  sort(e + 1, e + tot + 1, cmp);
  for (int i = (1); i <= (n); i++) f[i][i][0] = 1;
  for (int i = (1); i <= (tot); i++)
    for (int j = (1); j <= (n); j++)
      for (int k = (1); k <= (5); k++) {
        if (!f[j][e[i].x][k - 1]) continue;
        f[j][e[i].y][k] += f[j][e[i].x][k - 1];
      }
  for (int i = (1); i <= (n); i++) ans += f[i][i][5];
  printf("%lld\n", ans);
  return 0;
}
