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 double eps = 1e-10;
const int MX = 405, INF = 100000000;
int n;
int x[MX], y[MX];
long long int dp[6][MX][MX];
pair<double, pair<int, int> > e[MX * MX];
int ei;
int main() {
  int te;
  te = 1;
  while (te--) {
    scanf("%d", &n);
    for (int i = 0; i < n; i++) scanf("%d%d", &x[i], &y[i]);
    for (int k = 0; k < 6; k++)
      for (int i = 0; i < n; i++)
        for (int j = 0; j < n; j++) dp[k][i][j] = 0;
    ei = 0;
    for (int i = 0; i < n; i++) {
      dp[0][i][i] = 1;
      for (int j = 0; j < n; j++) {
        if (i == j) continue;
        e[ei++] = pair<double, pair<int, int> >(atan2(x[j] - x[i], y[j] - y[i]),
                                                pair<int, int>(i, j));
      }
    }
    sort(e, e + ei);
    int a, b;
    for (int i = 0; i < ei; i++) {
      a = e[i].second.first;
      b = e[i].second.second;
      for (int k = 0; k < 5; k++)
        for (int j = 0; j < n; j++) dp[k + 1][j][b] += dp[k][j][a];
    }
    long long int ans = 0;
    for (int i = 0; i < n; i++) ans += dp[5][i][i];
    printf("%lld\n", ans);
  }
  return 0;
}
