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;
struct point {
  int x, y;
  point(int x = 0, int y = 0) : x(x), y(y) {}
  point operator-(point a) { return point(x - a.x, y - a.y); }
  point operator+(point a) { return point(x + a.x, y + a.y); }
  point operator*(int a) { return point(x * a, y * a); }
  long long operator^(point a) {
    return (long long)x * a.y - (long long)y * a.x;
  }
  long long operator*(point a) {
    return (long long)x * a.x + (long long)y * a.y;
  }
  double nor() { return sqrt(*this * *this); }
  bool operator<(const point &a) const {
    if (x != a.x) return x < a.x;
    return y < a.y;
  }
  bool operator>(const point &a) const {
    if (x != a.x) return x > a.x;
    return y > a.y;
  }
  bool operator==(const point &a) const {
    if (x != a.x) return 0;
    return y == a.y;
  }
  void input() { scanf("%d %d", &x, &y); }
  void print() { printf("%d %d\n", x, y); }
};
point p[310];
bitset<310> bit[310][310];
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) p[i].input();
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++) {
      if (j == i) continue;
      for (int k = 1; k <= n; k++) {
        if (k == i || k == j) continue;
        if ((p[j] - p[i] ^ p[k] - p[i]) > 0) bit[i][j].set(k);
      }
    }
  }
  long long cnt1 = 0, cnt2 = 0;
  for (int i = 1; i <= n; i++) {
    for (int j = i + 1; j <= n; j++) {
      for (int k = j + 1; k <= n; k++) {
        int a = i, b = j, c = k;
        if ((p[c] - p[b] ^ p[a] - p[b]) < 0) swap(b, c);
        int t = (bit[a][b] & bit[b][c] & bit[c][a]).count();
        int t1;
        cnt1 += t * (t - 1) / 2;
        t1 = (bit[a][b] & bit[b][c]).count();
        cnt2 += t * (t1 - t);
        t1 = (bit[b][c] & bit[c][a]).count();
        cnt2 += t * (t1 - t);
        t1 = (bit[c][a] & bit[a][b]).count();
        cnt2 += t * (t1 - t);
      }
    }
  }
  long long ans = (long long)n * (n - 1) * (n - 2) * (n - 3) * (n - 4) / 120 -
                  cnt1 - cnt2 / 2;
  printf("%I64d\n", ans);
  return 0;
}
