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 ll = long long;
using ld = double;
const ll MODBASE = 1000000007LL;
const int MAXN = 310;
const int MAXM = 1000;
const int MAXK = 16;
const int MAXQ = 200010;
int n, x[MAXN], y[MAXN];
ll dp[MAXN][MAXN][6];
vector<pair<ld, pair<int, int> > > v;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(nullptr);
  cin >> n;
  for (int i = (1); i <= (n); i++) cin >> x[i] >> y[i];
  for (int i = (1); i <= (n); i++)
    for (int j = (1); j <= (n); j++)
      if (i != j) {
        int X = x[j] - x[i];
        int Y = y[j] - y[i];
        v.push_back(
            pair<ld, pair<int, int> >(atan2(Y, X), pair<int, int>(i, j)));
      }
  sort(v.begin(), v.end());
  for (int i = (1); i <= (n); i++)
    for (int j = (1); j <= (n); j++) {
      for (int k = (0); k <= (5); k++) dp[i][j][k] = 0;
      if (i == j) dp[i][j][0] = 1;
    }
  for (pair<ld, pair<int, int> > w : v) {
    int u = w.second.first, v = w.second.second;
    for (int i = (1); i <= (n); i++)
      for (int j = (0); j <= (4); j++) dp[i][v][j + 1] += dp[i][u][j];
  }
  ll res = 0;
  for (int i = (1); i <= (n); i++) res += dp[i][i][5];
  cout << res;
  return 0;
}
