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>
inline int read() {
  int res = 0;
  bool bo = 0;
  char c;
  while (((c = getchar()) < '0' || c > '9') && c != '-')
    ;
  if (c == '-')
    bo = 1;
  else
    res = c - 48;
  while ((c = getchar()) >= '0' && c <= '9')
    res = (res << 3) + (res << 1) + (c - 48);
  return bo ? ~res + 1 : res;
}
const int N = 305;
struct point {
  int x, y;
  friend inline point operator-(point a, point b) {
    return (point){b.x - a.x, b.y - a.y};
  }
  friend inline long long operator*(point a, point b) {
    return 1ll * a.x * b.y - 1ll * a.y * b.x;
  }
  inline double theta() { return atan2(y, x); }
} a[N];
int n, I, pt[N][N], tot, p[N], cnt[N][N], pts[N << 1];
bool vis[N];
long long ans;
inline bool comp(int x, int y) {
  return (a[I] - a[x]).theta() < (a[I] - a[y]).theta();
}
int main() {
  n = read();
  for (int i = 1; i <= n; i++) a[i].x = read(), a[i].y = read();
  for (I = 1; I <= n; I++) {
    for (int i = 1; i < n; i++) pt[I][i] = i < I ? i : i + 1;
    std::sort(pt[I] + 1, pt[I] + n, comp);
  }
  for (int i = 1; i <= n; i++) {
    memset(vis, 0, sizeof(vis));
    tot = 0;
    for (int j = 1; j < n; j++)
      if (a[pt[i][j]].y > a[i].y ||
          (a[pt[i][j]].y == a[i].y && a[pt[i][j]].x > a[i].x))
        vis[p[++tot] = pt[i][j]] = 1;
    for (int j = 1; j <= tot; j++) {
      int u = p[j], kt = n - 1 << 1, c = 0;
      for (int k = 1; k < n; k++) pts[k] = pts[k + n - 1] = pt[u][k];
      for (int k = (n - 1 << 1); k >= 1; k--) {
        if (vis[pts[k]] && (a[i] - a[pts[k]]) * (a[i] - a[u]) > 0) c++;
        while (kt == k + n ||
               (kt > k && (a[u] - a[pts[k]]) * (a[u] - a[pts[kt]]) < 0)) {
          if (vis[pts[kt]] && (a[i] - a[pts[kt]]) * (a[i] - a[u]) > 0) c--;
          kt--;
        }
        if (k < n && vis[pts[k]] && (a[i] - a[pts[k]]) * (a[i] - a[u]) < 0)
          cnt[u][pts[k]] = c;
      }
    }
    for (int j = 1; j <= tot; j++) {
      int u = p[j], kt = 1, c = 0;
      for (int k = 1; k < n; k++) pts[k] = pts[k + n - 1] = pt[u][k];
      for (int k = 1; k <= (n - 1 << 1); k++) {
        if (vis[pts[k]] && (a[i] - a[pts[k]]) * (a[i] - a[u]) < 0) c++;
        while (kt == k - n ||
               (kt < k && (a[u] - a[pts[k]]) * (a[u] - a[pts[kt]]) > 0)) {
          if (vis[pts[kt]] && (a[i] - a[pts[kt]]) * (a[i] - a[u]) < 0) c--;
          kt++;
        }
        if (k >= n && vis[pts[k]] && (a[i] - a[pts[k]]) * (a[i] - a[u]) > 0)
          cnt[u][pts[k]] = c;
      }
    }
    for (int j = 1; j <= tot; j++)
      for (int k = j + 1; k <= tot; k++)
        ans += cnt[p[j]][p[k]] * cnt[p[k]][p[j]];
  }
  return std::cout << ans << std::endl, 0;
}
