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;
inline int read() {
  int w = 1, s = 0;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') w = -1;
    ch = getchar();
  }
  while (isdigit(ch)) {
    s = s * 10 + ch - '0';
    ch = getchar();
  }
  return w * s;
}
struct node {
  int x, y;
} pt[330];
inline node operator+(node p, node q) { return (node){p.x + q.x, p.y + q.y}; }
inline node operator-(node p, node q) { return (node){p.x - q.x, p.y - q.y}; }
inline long long operator*(node p, node q) {
  return 1ll * p.x * q.y - 1ll * p.y * q.x;
}
struct line {
  int S, T;
  node o;
} l[1001010];
inline bool operator<(line p, line q) {
  if ((p.o * q.o) > 0) return 1;
  if ((p.o * q.o) < 0) return 0;
  if (p.S != q.S)
    return p.S < q.S;
  else
    return p.T < q.T;
}
int n, tot;
long long f[330][330][6];
int main() {
  n = read();
  for (register int i = 1; i <= n; ++i) {
    pt[i].x = read(), pt[i].y = read();
  }
  for (register int i = 1; i <= n; ++i) {
    for (register int j = 1; j <= n; ++j) {
      if (i == j) continue;
      l[++tot].S = i, l[tot].T = j, l[tot].o = pt[j] - pt[i];
    }
  }
  sort(l + 1, l + tot + 1);
  for (register int i = 1; i <= tot; ++i) {
    int x = l[i].S, y = l[i].T;
    f[x][y][1] += 1;
    for (register int j = 1; j <= n; ++j) {
      for (register int k = 1; k <= 4; ++k) f[j][y][k + 1] += f[j][x][k];
    }
  }
  long long Ans = 0;
  for (register int i = 1; i <= n; ++i) Ans += f[i][i][5];
  cout << Ans;
  return 0;
}
