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 {
  long long x, y;
  node(long long X = 0, long long Y = 0) : x(X), y(Y) {}
  node operator+(const node a) const { return node(x + a.x, y + a.y); }
  node operator-(const node a) const { return node(x - a.x, y - a.y); }
  node operator*(const long long a) const { return node(x * a, y * a); }
  long long operator*(const node a) const { return x * a.x + y * a.y; }
  long long operator^(const node a) const { return x * a.y - y * a.x; }
  bool operator<(const node a) const { return x == a.x ? y < a.y : x < a.x; }
} p[310], s[310];
double slope(const node a) { return atan2(a.y, a.x); }
double dis(const node a) { return sqrt(a.x * a.x + a.y * a.y); }
long long cross(const node a, const node o, const node b) {
  return (a - o) ^ (b - o);
}
bool cmp(const node a, const node b) { return slope(a) > slope(b); }
long long f[5][310][310];
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%lld%lld", &p[i].x, &p[i].y);
  sort(p + 1, p + n + 1);
  long long ans = 0;
  for (int u = 1; u <= n; u++) {
    int pn = 0;
    memset(f, 0, sizeof(f));
    for (int i = u; i <= n; i++) s[++pn] = p[i] - p[u];
    sort(s + 2, s + pn + 1, cmp);
    for (int i = 2; i <= pn; i++) f[1][i][1] = 1;
    for (int l = 1; l <= 3; l++)
      for (int i = l + 1; i <= pn; i++)
        for (int j = 1; j < i; j++)
          if (f[l][i][j])
            for (int k = i + 1; k <= pn; k++)
              if (cross(s[i], s[j], s[k]) < 0) f[l + 1][k][i] += f[l][i][j];
    for (int i = 1; i <= pn; i++)
      for (int j = 1; j < i; j++)
        if (cross(s[i], s[j], s[1]) < 0) ans += f[4][i][j];
  }
  printf("%lld", ans);
  return 0;
}
