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;
pair<int, int> pos[305];
long long int dp[305][7];
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    int x, y;
    scanf("%d %d", &x, &y);
    pos[i] = pair<int, int>(x, y);
  }
  sort(pos, pos + n);
  long long int ret = 0;
  for (int i = 0; i < n; i++) {
    vector<pair<double, pair<int, int> > > vx;
    for (int j = i; j < n; j++) {
      for (int k = i; k < n; k++) {
        if (j == k) continue;
        pair<int, int> p = pos[k];
        p.first -= pos[j].first;
        p.second -= pos[j].second;
        double d = atan2(p.first, -p.second);
        if (d < 1e-10) d += 2.0 * 3.1415926535897932384626;
        vx.push_back(pair<double, pair<int, int> >(d, pair<int, int>(j, k)));
      }
    }
    sort(vx.begin(), vx.end());
    memset(dp, 0, sizeof(dp));
    dp[i][0] = 1;
    for (int j = 0; j < vx.size(); j++) {
      pair<int, int> p = vx[j].second;
      for (int t = 0; t + 1 < 7; t++) {
        dp[p.second][t + 1] += dp[p.first][t];
      }
    }
    ret += dp[i][5];
  }
  printf("%lld\n", ret);
  return 0;
}
