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;
template <class t>
inline t read(t &x) {
  bool f = 0;
  char c = getchar();
  x = 0;
  while (!isdigit(c)) f |= c == '-', c = getchar();
  while (isdigit(c)) x = (x << 1) + (x << 3) + (c ^ 48), c = getchar();
  if (f) x = -x;
  return x;
}
template <class t>
inline void write(t x) {
  if (x < 0)
    putchar('-'), write(-x);
  else {
    if (x > 9) write(x / 10);
    putchar('0' + x % 10);
  }
}
const int N = 305;
int n;
pair<int, int> a[N];
bitset<N> b[N][N];
long long all, C3, C4;
pair<int, int> operator-(pair<int, int> first, pair<int, int> second) {
  return {first.first - second.first, first.second - second.second};
}
long long cj(pair<int, int> a, pair<int, int> b) {
  return 1ll * a.first * b.second - 1ll * a.second * b.first;
}
signed main() {
  read(n);
  all = 1ll * n * (n - 1) / 2 * (n - 2) / 3 * (n - 3) / 4 * (n - 4) / 5;
  for (int i = 0; i < n; i++) read(a[i].first), read(a[i].second);
  for (int i = 0; i < n; i++)
    for (int j = i + 1; j < n; j++)
      for (int k = j + 1; k < n; k++) {
        if (cj(a[j] - a[i], a[k] - a[i]) >= 0)
          b[i][j][k] = b[j][k][i] = b[k][i][j] = 1;
        else
          b[i][k][j] = b[j][i][k] = b[k][j][i] = 1;
      }
  for (int i = 0; i < n; i++)
    for (int j = i + 1; j < n; j++)
      for (int k = j + 1; k < n; k++) {
        int w = (b[i][j][k] ? b[i][j] & b[j][k] & b[k][i]
                            : b[i][k] & b[k][j] & b[j][i])
                    .count();
        C3 += w * (w - 1) / 2;
        C4 += w;
      }
  C4 = C4 * (n - 4) / 2 - C3 * 2;
  write(all - C3 - C4);
}
