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 long long maxn = 305;
long long po[maxn][maxn];
long long kiek[maxn];
pair<long long, long long> a[maxn];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  long long n;
  cin >> n;
  for (long long i = 0; i < n; ++i) cin >> a[i].first >> a[i].second;
  sort(a, a + n);
  for (long long i = 0; i < n; ++i) {
    for (long long j = i + 1; j < n; ++j) {
      for (long long k = i + 1; k < j; ++k) {
        if (a[k].first == a[i].first || a[k].first == a[j].first) continue;
        if ((a[k].first - a[i].first) * (a[j].second - a[i].second) -
                (a[k].second - a[i].second) * (a[j].first - a[i].first) >
            0)
          ++po[i][j];
      }
    }
  }
  for (int i = 0; i < n; ++i) {
    for (int j = i + 1; j < n; ++j) {
      for (int k = j + 1; k < n; ++k) {
        int kur = abs(po[i][k] - po[i][j] - po[j][k]);
        if (a[j].first != a[i].first && a[j].first != a[k].first) {
          if ((a[j].first - a[i].first) * (a[k].second - a[i].second) -
                  (a[j].second - a[i].second) * (a[k].first - a[i].first) >
              0)
            --kur;
        }
        ++kiek[kur];
      }
    }
  }
  long long ans = n * (n - 1) * (n - 2) * (n - 3) * (n - 4) / 120;
  long long r = 0;
  for (int i = 0; i < maxn; ++i) {
    r -= i * kiek[i] * (n - 3 - i);
  }
  ans += r / 2;
  cout << ans;
  return 0;
}
