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 mxn = 300, mxm = mxn * mxn, k = 5;
int n, m;
long long a[mxn], b[mxn];
pair<long long, long long> first[mxm];
long long dp[mxn][mxn][k + 1];
pair<long long, long long> ln(pair<long long, long long> x) {
  return {a[x.second] - a[x.first], b[x.second] - b[x.first]};
}
int hf(pair<long long, long long> x) { return (x.first ?: x.second) < 0; }
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> a[i] >> b[i];
    dp[i][i][0] = 1;
    for (int j = 0; j < i; j++) first[m++] = {i, j}, first[m++] = {j, i};
  }
  sort(first, first + m,
       [&](pair<long long, long long> x, pair<long long, long long> y) {
         pair<long long, long long> p = ln(x), q = ln(y);
         return hf(p) == hf(q) ? p.first * q.second > p.second * q.first
                               : hf(p) < hf(q);
       });
  for (int i = 0; i < m; i++)
    for (int j = 0; j < n; j++)
      for (int l = 1; l <= k; l++) {
        dp[j][first[i].second][l] += dp[j][first[i].first][l - 1];
      }
  long long ret = 0;
  for (int i = 0; i < n; i++) ret += dp[i][i][k];
  cout << ret << '\n';
  return 0;
}
