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 {
  long long x, y;
};
struct vec {
  long long x, y;
  vec(point p1, point p2) {
    x = p2.x - p1.x;
    y = p2.y - p1.y;
  }
  bool operator<(vec o) const { return atan2(y, x) < atan2(o.y, o.x); }
};
int n;
point P[310];
long long dp[310][310][6];
int main() {
  while (scanf("%d", &n) > 0) {
    for (int i = 0; i < n; i++) {
      scanf("%lld%lld", &P[i].x, &P[i].y);
    }
    vector<pair<vec, pair<int, int> > > seg;
    for (int i = 0; i < n; i++) {
      for (int j = 0; j < n; j++) {
        if (i == j) continue;
        seg.push_back(make_pair(vec(P[i], P[j]), pair<int, int>(i, j)));
      }
    }
    sort(seg.begin(), seg.end());
    memset(dp, 0, sizeof dp);
    for (int k = 0; k < seg.size(); k++) {
      int a = seg[k].second.first, b = seg[k].second.second;
      dp[a][b][1]++;
      for (int s = 1; s < 5; s++) {
        for (int i = 0; i < n; i++) {
          dp[i][b][s + 1] += dp[i][a][s];
        }
      }
    }
    long long ans = 0;
    for (int i = 0; i < n; i++) ans += dp[i][i][5];
    printf("%lld\n", ans);
  }
}
