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 = 303;
int n;
int x[maxn], y[maxn];
int f[6][maxn][maxn];
struct Edge {
  int u, v;
  double Ang;
} e[maxn * maxn];
int tot_Edge;
inline bool cmp(const Edge &A, const Edge &B) { return A.Ang < B.Ang; }
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[++tot_Edge] = (Edge){i, j, atan2(y[j] - y[i], x[j] - x[i])};
      }
  }
  stable_sort(e + 1, e + 1 + tot_Edge, cmp);
  for (int id = 1; id <= tot_Edge; id++) {
    for (int i = 5; i > 1; i--) {
      for (int u = 1; u <= n; u++) {
        f[i][u][e[id].v] += f[i - 1][u][e[id].u];
      }
    }
    f[1][e[id].u][e[id].v] = 1;
  }
  long long Ans = 0;
  for (int u = 1; u <= n; u++) Ans += f[5][u][u];
  printf("%lld\n", Ans);
  return 0;
}
