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;
struct point {
  long long x, y;
  int ck, num;
  point operator-(const point &p) const { return {x - p.x, y - p.y, ck, num}; }
  bool operator<(const point &p) const {
    if ((x > 0 || (x == 0 && y > 0)) != (p.x > 0 || (p.x == 0 && p.y > 0)))
      return (x > 0 || (x == 0 && y > 0));
    return y * p.x < p.y * x;
  }
} w[310], P[610];
int n, A[310][310][310], B[310][310][310];
long long ccw(point a, point b, point c) {
  return (b.x - a.x) * (c.y - a.y) - (b.y - a.y) * (c.x - a.x);
}
int main() {
  int i, j, k;
  scanf("%d", &n);
  for (i = 1; i <= n; i++) {
    scanf("%lld%lld", &w[i].x, &w[i].y);
  }
  for (i = 1; i <= n; i++) {
    int cnt = 0;
    for (j = 1; j <= n; j++) {
      if (j != i) {
        P[cnt] = w[j] - w[i];
        P[cnt].num = j, P[cnt].ck = 0;
        cnt++;
        P[cnt] = w[i] - w[j];
        P[cnt].num = j, P[cnt].ck = 1;
        cnt++;
      }
    }
    sort(P, P + cnt);
    for (j = 0; j < cnt; j++) {
      int c = 0;
      for (k = 0; k < cnt; k++) {
        int t = (j + k) % cnt;
        if (!P[t].ck) c++;
        if (P[t].ck == P[j].ck) {
          if (P[j].ck == 0)
            A[i][P[j].num][P[t].num] = c;
          else
            B[i][P[j].num][P[t].num] = c;
        }
      }
    }
  }
  long long r1 = 0, r2 = 0;
  int s = 0;
  for (i = 1; i <= n; i++) {
    for (j = i + 1; j <= n; j++) {
      for (k = i + 1; k <= n; k++) {
        if (i == j || i == k || j == k) continue;
        if (ccw(w[i], w[j], w[k]) <= 0) continue;
        int t = A[i][j][k] + B[i][j][k] + A[j][k][i] + B[j][k][i] + A[k][i][j] +
                B[k][i][j] - n - 3;
        t /= 2;
        r1 += t * (t - 1) / 2;
        r2 += t * (A[i][j][k] - t - 2);
        r2 += t * (A[j][k][i] - t - 2);
        r2 += t * (A[k][i][j] - t - 2);
      }
    }
  }
  r2 /= 2;
  long long res = 1ll * n * (n - 1) * (n - 2) * (n - 3) * (n - 4) / 120;
  printf("%lld\n", res - r1 - r2);
}
