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 = 305;
struct node {
  int x, y;
} a[N], e[N * N];
int n, i, j, k, cnt;
long long f[N][N][10], ans;
node operator-(node a, node b) { return (node){a.x - b.x, a.y - b.y}; }
long long mult(node a, node b) { return 1ll * a.x * b.y - 1ll * a.y * b.x; }
int xx(node a) {
  if (a.x > 0) {
    if (a.y > 0)
      return 1;
    else
      return 4;
  } else {
    if (a.y > 0)
      return 2;
    else
      return 3;
  }
}
bool cmp(node p, node q) {
  node g = a[p.y] - a[p.x], h = a[q.y] - a[q.x];
  int xx1 = xx(g), xx2 = xx(h);
  if (xx1 != xx2) return xx1 < xx2;
  return mult(g, h) > 0;
}
int main() {
  scanf("%d", &n);
  for (i = 1; i <= n; ++i) scanf("%d%d", &a[i].x, &a[i].y);
  for (i = 1; i <= n; ++i)
    for (j = 1; j <= n; ++j)
      if (i != j) e[++cnt] = (node){i, j};
  sort(e + 1, e + 1 + cnt, cmp);
  for (i = 1; i <= cnt; ++i) {
    f[e[i].x][e[i].y][1]++;
    for (j = 1; j <= n; ++j)
      for (k = 1; k <= 4; ++k) {
        f[j][e[i].y][k + 1] += f[j][e[i].x][k] * f[e[i].x][e[i].y][1];
      }
  }
  for (i = 1; i <= n; ++i) ans += f[i][i][5];
  printf("%lld", ans);
  return 0;
}
