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 int maxn = 312;
pair<double, pair<int, int> > e[maxn * maxn];
int x[maxn], y[maxn], dp[maxn][maxn][5];
int n, m;
long long ans;
int main() {
  scanf("%d", &n);
  for (int(i) = (1); (i) <= (n); (i)++) scanf("%d%d", x + i, y + i);
  for (int(i) = (1); (i) <= (n); (i)++)
    for (int(j) = (1); (j) <= (n); (j)++)
      if (i != j) e[++m] = {atan2(y[j] - y[i], x[j] - x[i]), make_pair(i, j)};
  sort(e + 1, e + m + 1);
  for (int(i) = (1); (i) <= (m); (i)++) {
    int u = e[i].second.first, v = e[i].second.second;
    dp[u][v][0] = 1;
    for (int(j) = (1); (j) <= (n); (j)++)
      for (int(k) = (0); (k) <= ((4) - 1); (k)++)
        dp[j][v][k + 1] += dp[j][u][k];
  }
  for (int(i) = (1); (i) <= (n); (i)++) ans += dp[i][i][4];
  printf("%lld", ans);
}
