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 NN = 301;
bitset<NN> a, b, c, in, all, val[NN][NN];
int cnt[NN][NN];
pair<int, int> p[NN];
pair<int, int> operator-(const pair<int, int> &a, const pair<int, int> &b) {
  return pair<int, int>(a.first - b.first, a.second - b.second);
}
long long operator^(const pair<int, int> &a, const pair<int, int> &b) {
  return (long long)a.first * b.second - (long long)a.second * b.first;
}
int sgn(long long a) { return (a > 0) - (a < 0); }
int main() {
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) scanf("%d%d", &p[i].first, &p[i].second);
  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 (k != i && k != j) {
            if (sgn(p[j] - p[i] ^ p[k] - p[i]) > 0) val[i][j].set(k);
          }
      }
  }
  for (int i = 0; i < n; i++)
    for (int j = 0; j < n; j++) cnt[i][j] = val[i][j].count();
  for (int i = 0; i < n; i++) all[i] = 1;
  long long B = 0, C = 0;
  for (int i = 0; i < n; i++)
    for (int j = i + 1; j < n; j++)
      for (int k = j + 1; k < n; k++) {
        int u = i, v = j, w = k;
        if (sgn(p[w] - p[u] ^ p[v] - p[u]) < 0) swap(v, w);
        b = val[u][v] & val[v][w];
        c = val[v][w] & val[w][u];
        a = val[w][u] & val[u][v];
        in = all ^ val[u][v] ^ val[v][w] ^ val[w][u] ^ a ^ b ^ c;
        int d = in.count() - 3, aa = a.count(), bb = b.count(), cc = c.count();
        B += d * (d - 1) >> 1;
        C += d * (cnt[u][v] + cnt[v][w] + cnt[w][u] - (aa + bb + cc << 1));
      }
  cout << (long long)n * (n - 1) / 2 * (n - 2) / 3 * (n - 3) / 4 * (n - 4) / 5 -
              B - C / 2
       << endl;
  return 0;
}
