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 vec {
  long long x, y;
  vec operator+(const vec &tmp) const { return (vec){x + tmp.x, y + tmp.y}; }
  vec operator-(const vec &tmp) const { return (vec){x - tmp.x, y - tmp.y}; }
} a[305];
long long n;
bitset<305> e[305][305];
long long sig(long long x) {
  if (x > 0)
    return 1;
  else if (x < 0)
    return -1;
  else
    return 0;
  assert(0);
}
long long cross(vec a, vec b, vec c) {
  b = b - a;
  c = c - a;
  return b.x * c.y - b.y * c.x;
}
signed main() {
  scanf("%lld", &n);
  for (long long i = 1; i <= n; i++) scanf("%lld%lld", &a[i].x, &a[i].y);
  for (long long i = 1; i <= n; i++)
    for (long long j = i + 1; j <= n; j++)
      for (long long k = j + 1; k <= n; k++)
        if (sig(cross(a[i], a[j], a[k])) == -1)
          e[i][k].set(j), e[k][j].set(i), e[j][i].set(k);
        else
          e[i][j].set(k), e[j][k].set(i), e[k][i].set(j);
  long long res3 = 0, res4 = 0;
  for (long long i = 1; i <= n; i++)
    for (long long j = i + 1; j <= n; j++)
      for (long long k = j + 1; k <= n; k++) {
        long long tmp = (e[i][j].test(k) ? (e[i][j] & e[j][k] & e[k][i])
                                         : (e[i][k] & e[k][j] & e[j][i]))
                            .count();
        res3 += tmp * (tmp - 1) / 2;
        res4 += tmp;
      }
  res4 = res4 * (n - 4) / 2 - res3 * 2;
  printf("%lld\n", n * (n - 1) / 2 * (n - 2) / 3 * (n - 3) / 4 * (n - 4) / 5 -
                       res3 - res4);
  return 0;
}
