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>
struct Point {
  int x, y;
  Point(int x, int y) : x(x), y(y) {}
  Point() : Point(0, 0) {}
  void read() { scanf("%d %d", &x, &y); }
  struct Point operator-(struct Point p) const {
    return Point(x - p.x, y - p.y);
  }
  double arg() { return atan2(y, x); }
} ps[305];
int64_t dp[305][6];
int main() {
  int N;
  scanf("%d", &N);
  for (int i = 0; i < N; i++) {
    ps[i].read();
  }
  std::vector<std::pair<double, std::pair<int, int> > > edges;
  for (int i = 0; i < N; i++) {
    for (int j = 0; j < N; j++) {
      if (i == j) continue;
      edges.emplace_back((ps[j] - ps[i]).arg(), std::make_pair(i, j));
    }
  }
  std::sort(edges.begin(), edges.end());
  int64_t ans = 0;
  for (int i = 0; i < N; i++) {
    for (int j = 0; j < N; j++) {
      for (int k = 0; k < 6; k++) {
        dp[j][k] = 0;
      }
    }
    dp[i][0] = 1;
    for (auto e : edges) {
      for (int k = 0; k < 5; k++) {
        dp[e.second.second][k + 1] += dp[e.second.first][k];
      }
    }
    ans += dp[i][5];
  }
  printf("%I64d\n", ans);
  return 0;
}
