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 node {
  int x, y;
  double jd;
  node() {}
  node(int x, int y, double jd) : x(x), y(y), jd(jd) {}
};
bool cmp(const node &a, const node &b) { return a.jd < b.jd; }
const int maxn = 333;
double dp[maxn][maxn][6];
node p[maxn * maxn];
int x[maxn];
int y[maxn];
int main() {
  int n;
  memset(dp, 0, sizeof(dp));
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%d%d", &x[i], &y[i]);
  }
  int cnt = 0;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) {
      if (i == j) continue;
      p[cnt] = node(i, j, atan2(y[i] - y[j], x[i] - x[j]));
      cnt++;
    }
  }
  sort(p, p + cnt, cmp);
  for (int i = 0; i < cnt; i++) {
    dp[p[i].x][p[i].y][1] = 1;
    for (int j = 0; j < n; j++) {
      for (int k = 2; k <= 5; k++) {
        dp[j][p[i].y][k] += dp[j][p[i].x][k - 1];
      }
    }
  }
  long long ans = 0;
  for (int i = 0; i < n; i++) {
    ans += dp[i][i][5];
  }
  cout << ans;
  return 0;
}
