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 N = 300 + 5;
vector<pair<double, pair<int, int>>> e;
int x[N], y[N];
long long dp[N][N][5];
int main() {
  int n;
  cin >> n;
  for (int i = 1; i <= n; ++i) cin >> x[i] >> y[i];
  for (int i = 1; i <= n; ++i)
    for (int j = 1; j <= n; ++j)
      if (j != i) e.push_back({atan2(x[j] - x[i], y[j] - y[i]), {i, j}});
  sort(e.begin(), e.end());
  for (auto p : e) {
    int x, y;
    tie(x, y) = p.second;
    ++dp[x][y][0];
    for (int i = 1; i <= n; ++i)
      for (int j = 0; j <= 3; ++j) dp[i][y][j + 1] += dp[i][x][j];
  }
  long long ans = 0;
  for (int i = 1; i <= n; ++i) ans += dp[i][i][4];
  cout << ans;
}
