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;
template <class A, class B>
inline bool mina(A &first, B second) {
  return (first > second) ? (first = second, 1) : 0;
}
template <class A, class B>
inline bool maxa(A &first, B second) {
  return (first < second) ? (first = second, 1) : 0;
}
const int MAXN = 305;
int N;
pair<int, int> in[MAXN];
vector<pair<double, pair<int, int> > > lines;
long long dp[MAXN][MAXN][6];
int main() {
  scanf("%d", &N);
  for (int i = 0; (i) < (N); ++(i)) {
    scanf("%d %d", &in[i].first, &in[i].second);
  }
  for (int i = 0; (i) < (N); ++(i)) {
    for (int j = 0; (j) < (N); ++(j)) {
      if (i != j) {
        int dx = in[j].first - in[i].first;
        int dy = in[j].second - in[i].second;
        lines.push_back(
            pair<double, pair<int, int> >(atan2(dy, dx), pair<int, int>(i, j)));
      }
    }
  }
  sort(lines.begin(), lines.end());
  for (int i = 0; (i) < (N); ++(i)) {
    dp[i][i][0] = 1;
  }
  for (int k = 0; (k) < (((int)(lines).size())); ++(k)) {
    int a = lines[k].second.first;
    int b = lines[k].second.second;
    for (int i = 0; (i) < (N); ++(i)) {
      for (int j = 0; (j) < (5); ++(j)) {
        dp[i][b][j + 1] += dp[i][a][j];
      }
    }
  }
  long long ans = 0;
  for (int i = 0; (i) < (N); ++(i)) {
    ans += dp[i][i][5];
  }
  cout << ans << endl;
  return 0;
}
