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;
char getc() {
  char c = getchar();
  while ((c < 'A' || c > 'Z') && (c < 'a' || c > 'z') && (c < '0' || c > '9'))
    c = getchar();
  return c;
}
int gcd(int n, int m) { return m == 0 ? n : gcd(m, n % m); }
int read() {
  int x = 0, f = 1;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9')
    x = (x << 1) + (x << 3) + (c ^ 48), c = getchar();
  return x * f;
}
int n, m, down_cnt[310], up_cnt[310];
long long ans;
double down[310][310], up[310][310];
struct point {
  int x, y;
  point operator+(const point& a) const { return (point){x + a.x, y + a.y}; }
  point operator-(const point& a) const { return (point){x - a.x, y - a.y}; }
  long long operator*(const point& a) const {
    return 1ll * x * a.y - 1ll * y * a.x;
  }
} a[310], b[310], O;
double alpha(point x) { return atan2(x.y, x.x); }
bool cmp1(const point& a, const point& b) { return a.y > b.y; }
bool cmp2(const point& a, const point& b) { return alpha(a) < alpha(b); }
bool onright(point a, point b, point c) { return (c - a) * (b - a) > 0; }
signed main() {
  n = read();
  for (int i = 1; i <= n; i++) a[i].x = read(), a[i].y = read();
  sort(a + 1, a + n + 1, cmp1);
  for (int i = 1; i <= n; i++) {
    m = 0;
    for (int j = i + 1; j <= n; j++) b[++m] = a[j] - a[i];
    sort(b + 1, b + m + 1, cmp2);
    for (int j = 1; j <= m; j++) {
      down_cnt[j] = 0;
      for (int k = 1; k < j; k++)
        if (onright(O, b[j], b[k])) down[j][++down_cnt[j]] = alpha(b[j] - b[k]);
      sort(down[j] + 1, down[j] + down_cnt[j] + 1);
      up_cnt[j] = 0;
      for (int k = j + 1; k <= m; k++)
        if (onright(O, b[k], b[j])) up[j][++up_cnt[j]] = alpha(b[k] - b[j]);
      sort(up[j] + 1, up[j] + up_cnt[j] + 1);
    }
    for (int j = 1; j <= m; j++)
      for (int k = j + 1; k <= m; k++) {
        int x = lower_bound(down[j] + 1, down[j] + down_cnt[j] + 1,
                            alpha(b[k] - b[j])) -
                down[j] - 1;
        int y =
            up_cnt[k] -
            (lower_bound(up[k] + 1, up[k] + up_cnt[k] + 1, alpha(b[k] - b[j])) -
             up[k]) +
            1;
        ans += 1ll * x * y;
      }
  }
  cout << ans;
  return 0;
}
