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 = 320;
struct point {
  long long x, y;
  point() {}
  point(long long a, long long b) : x(a), y(b){};
  point operator-(point a) { return point(x - a.x, y - a.y); }
  bool operator<(point a) const {
    if (x == a.x) return y < a.y;
    return x < a.x;
  }
} dt[maxn];
int n;
bitset<maxn> left_[maxn][maxn];
long long xmul(point a, point b) { return a.x * b.y - a.y * b.x; }
void init() {
  sort(dt + 1, dt + n + 1);
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++) {
      for (int k = 1; k <= n; k++) {
        if (xmul(dt[j] - dt[i], dt[k] - dt[i]) > 0) left_[i][j][k] = 1;
      }
    }
  }
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%lld %lld", &dt[i].x, &dt[i].y);
  init();
  long long ans = (long long)n * (n - 1) * (n - 2) * (n - 3) * (n - 4) / 120,
            sum = 0;
  for (int i = 1; i <= n; i++) {
    for (int j = i + 1; j <= n; j++) {
      for (int k = i + 1; k <= n; k++) {
        if (!left_[i][k][j]) continue;
        long long cnt = (left_[i][k] & left_[k][j] & left_[j][i]).count();
        sum += cnt * (n - 4);
        ans += cnt * (cnt - 1) / 2;
      }
    }
  }
  ans -= sum / 2;
  cout << ans << endl;
}
