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;
long long lgput(long long a, long long b, long long mod) {
  long long ret = 1;
  while (b) {
    if (b & 1) ret = (ret * a) % mod;
    a = (a * a) % mod;
    b >>= 1;
  }
  return (ret % mod);
}
inline long long inv(long long x, long long MOD) {
  return lgput(x, MOD - 2, MOD);
}
const long double PI = acos(-1);
const long double eps = 1e-6;
const int MAXN = 302;
long long dp[MAXN][MAXN][6];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cout.precision(12);
  cout.fixed;
  int n;
  cin >> n;
  vector<pair<int, int> > points(n);
  for (int i = 0; i < n; ++i) {
    cin >> points[i].first >> points[i].second;
  }
  vector<pair<int, int> > order;
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < n; ++j)
      if (i != j) order.emplace_back(i, j);
  }
  sort(order.begin(), order.end(),
       [&](pair<int, int> &a, pair<int, int> &b) -> bool {
         return atan2(points[a.second].second - points[a.first].second,
                      points[a.second].first - points[a.first].first) <
                atan2(points[b.second].second - points[b.first].second,
                      points[b.second].first - points[b.first].first);
       });
  for (int i = 0; i < n; ++i) dp[i][i][0] = 1;
  for (auto &x : order) {
    for (int k = 1; k <= 5; ++k) {
      for (int i = 0; i < n; ++i) {
        dp[i][x.second][k] += dp[i][x.first][k - 1];
      }
    }
  }
  long long ans = 0;
  for (int i = 0; i < n; ++i) ans += dp[i][i][5];
  cout << ans << '\n';
  return 0;
}
