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;
int n, x[305], y[305];
vector<pair<double, pair<int, int> > > v;
long long dp[305][305][6];
void add(long long &a, long long b) { a += b; }
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d%d", &x[i], &y[i]);
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= n; j++) {
      if (i == j) continue;
      double ang = atan2(y[j] - y[i], x[j] - x[i]);
      v.push_back(make_pair(ang, pair<int, int>(i, j)));
    }
  sort(v.begin(), v.end());
  for (int i = 1; i <= n; i++) dp[i][i][0] = 1;
  for (int i = 0; i < (int)v.size(); i++)
    for (int j = 1; j <= n; j++)
      for (int k = 0; k < 5; k++)
        add(dp[j][v[i].second.second][k + 1], dp[j][v[i].second.first][k]);
  long long ans = 0;
  for (int i = 1; i <= n; i++) add(ans, dp[i][i][5]);
  printf("%lld\n", ans);
  return 0;
}
