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;
using cat = long long;
struct pt {
  cat x, y;
  pt operator-(const pt& P) const { return {x - P.x, y - P.y}; }
  bool operator<(const pt& P) const {
    if (x != P.x) return x < P.x;
    return y < P.y;
  }
};
cat vs(pt A, pt B) { return A.x * B.y - A.y * B.x; }
int main() {
  cin.sync_with_stdio(0);
  cin.tie(0);
  cout << fixed << setprecision(10);
  int N;
  cin >> N;
  vector<pt> P(N);
  for (int i = 0; i < N; i++) cin >> P[i].x >> P[i].y;
  vector<vector<int> > G(N * N);
  for (int i = 0; i < N; i++)
    for (int j = 0; j < N; j++)
      if (i != j)
        for (int k = 0; k < N; k++)
          if (i != k && j != k)
            if (vs(P[k] - P[i], P[j] - P[i]) > 0)
              G[i * N + j].push_back(j * N + k);
  vector<vector<int> > cnt_c(N, vector<int>(N, 0));
  for (int i = 1; i < N; i++)
    for (int j = 1; j < N; j++) {
      if (vs(P[i] - P[0], P[j] - P[0]) <= 0) continue;
      bool sgn_0 = (vs(P[0] - P[i], P[j] - P[i]) > 0);
      for (int k = 1; k < N; k++)
        if (i != k && j != k) {
          if (vs(P[i] - P[0], P[k] - P[0]) <= 0) continue;
          if (vs(P[k] - P[0], P[j] - P[0]) <= 0) continue;
          bool sgn_k = (vs(P[k] - P[i], P[j] - P[i]) > 0);
          if (sgn_k == sgn_0) cnt_c[i][j]++;
        }
      cnt_c[j][i] = cnt_c[i][j];
    }
  cat ans = 1LL * N * (N - 1) * (N - 2) * (N - 3) * (N - 4) / 60;
  for (int i = 1; i < N; i++)
    for (int j = 1; j < N; j++) {
      if (vs(P[i] - P[0], P[j] - P[0]) <= 0) continue;
      ans -= 1LL * cnt_c[i][j] * (N - 4);
      ans += 1LL * cnt_c[i][j] * (cnt_c[i][j] - 1);
      for (int k = 1; k < N; k++)
        if (i != k && j != k) {
          bool s1 = (vs(P[i] - P[0], P[k] - P[0]) > 0);
          bool s2 = (vs(P[k] - P[0], P[j] - P[0]) > 0);
          if (s1 != s2) continue;
          if (!s1)
            if (k > i || k > j) continue;
          int cnt;
          if (s1) {
            cnt = cnt_c[i][k] + cnt_c[j][k] - cnt_c[i][j];
            if (cnt < 0) cnt = -cnt - 1;
          } else
            cnt = 1 + cnt_c[i][k] + cnt_c[j][k] + cnt_c[i][j];
          ans -= 1LL * cnt * (N - 4);
          ans += 1LL * cnt * (cnt - 1);
        }
    }
  ans /= 2;
  cout << ans << "\n";
  return 0;
}
