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 pii = pair<int, int>;
const int MAXN = 310;
int N;
pii P[MAXN];
int dp[MAXN][MAXN][6];
int main() {
  scanf("%d", &N);
  for (int i = 0; i < N; i++) {
    scanf("%d %d", &P[i].first, &P[i].second);
  }
  vector<pair<double, pii>> vecs;
  for (int i = 0; i < N; i++) {
    for (int j = 0; j < N; j++) {
      if (i == j) continue;
      int dx = P[j].first - P[i].first;
      int dy = P[j].second - P[i].second;
      vecs.emplace_back(atan2(dy, dx), pii(i, j));
    }
  }
  sort(vecs.begin(), vecs.end());
  for (int i = 0; i < N; i++) dp[i][i][0] = 1;
  for (auto v : vecs) {
    int i, j;
    tie(i, j) = v.second;
    for (int k = 0; k < N; k++) {
      for (int c = 0; c < 5; c++) {
        dp[k][j][c + 1] += dp[k][i][c];
      }
    }
  }
  ll ans = 0;
  for (int i = 0; i < N; i++) ans += dp[i][i][5];
  printf("%lld\n", ans);
  return 0;
}
