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;
istream &operator>>(istream &is, complex<long long> &p) {
  int X, Y;
  is >> X >> Y;
  p = {X, Y};
  return is;
}
const int N = 305;
long long n;
complex<long long> p[N];
long long dp[N][N][5];
struct seg {
  int i, j;
  seg(int i, int j) : i(i), j(j) {}
};
bool upper(const complex<long long> p) {
  return p.imag() > 0 || (p.imag() == 0 && p.real() > 0);
}
bool cmp(const seg *a, const seg *b) {
  if (upper(p[a->j] - p[a->i]) != upper(p[b->j] - p[b->i]))
    return upper(p[a->j] - p[a->i]);
  return ((conj(p[a->j] - p[a->i]) * (p[b->j] - p[b->i])).imag()) > 0;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> p[i];
  }
  vector<seg *> v;
  for (int i = 0; i < n; i++) {
    for (int j = i + 1; j < n; j++) {
      v.push_back(new seg(i, j));
      v.push_back(new seg(j, i));
    }
  }
  sort(v.begin(), v.end(), cmp);
  for (int t = 0; t < (int)v.size(); t++) {
    int i = v[t]->i, j = v[t]->j;
    for (int k = 0; k < n; k++) {
      for (int l = 1; l < 5; l++) {
        dp[i][k][l + 1] += dp[j][k][l];
      }
    }
    dp[i][j][1] = 1;
  }
  long long ans = 0;
  for (int i = 0; i < n; i++) {
    ans += dp[i][i][5];
  }
  cout << ans << endl;
}
