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 maxn = 305;
long long dp[maxn][maxn][6];
int n, x[maxn], y[maxn];
vector<pair<double, pair<int, int>>> seg;
int main() {
  ios_base::sync_with_stdio(0);
  cin >> n;
  for (int i = 0; i < n; i++) cin >> x[i] >> y[i];
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) {
      for (int k = 0; k < 6; k++) dp[i][j][k] = 0;
      if (i == j) dp[i][j][0] = 1;
    }
  }
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) {
      if (i != j) {
        int xx = x[i] - x[j], yy = y[i] - y[j];
        seg.push_back({atan2(xx, yy), {i, j}});
      }
    }
  }
  sort(seg.begin(), seg.end());
  int m = seg.size();
  for (int cc = 0; cc < m; cc++) {
    int i = seg[cc].second.first, j = seg[cc].second.second;
    for (int k = 0; k < n; k++)
      for (int f = 0; f < 5; f++) dp[k][j][f + 1] += dp[k][i][f];
  }
  long long ret = 0;
  for (int i = 0; i < n; i++) ret += dp[i][i][5];
  cout << ret << endl;
  return 0;
}
