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;
template <class T>
inline void rd(T &x) {
  x = 0;
  char c = getchar();
  int f = 1;
  while (!isdigit(c)) {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (isdigit(c)) x = x * 10 - '0' + c, c = getchar();
  x *= f;
}
const int N = 310;
struct Point {
  long long x, y;
  Point(long long x = 0, long long y = 0) : x(x), y(y) {}
  friend Point operator-(Point A, Point B) {
    return Point(A.x - B.x, A.y - B.y);
  }
  friend long long operator^(Point A, Point B) { return A.x * B.y - A.y * B.x; }
  friend long long operator*(Point A, Point B) { return A.x * B.x + A.y * B.y; }
  friend bool operator<(Point A, Point B) {
    return A.x == B.x ? A.y < B.y : A.x < B.x;
  }
} p[N];
int sum[N][N], n;
int check(Point A, Point B, Point C) {
  if (C.x < A.x || C.x > B.x) return 0;
  return ((C - A) ^ (B - A)) > 0;
}
int Abs(int x) { return x > 0 ? x : -x; }
int main() {
  rd(n);
  for (int i = 1; i <= n; ++i) rd(p[i].x), rd(p[i].y);
  sort(p + 1, p + n + 1);
  for (int i = 1; i <= n; ++i)
    for (int j = i + 1; j <= n; ++j) {
      for (int k = 1; k <= n; ++k)
        if (k != i && k != j) sum[i][j] += check(p[i], p[j], p[k]);
    }
  long long ans1 = 0, ans2 = 0;
  for (int i = 1; i <= n; ++i)
    for (int j = i + 1; j <= n; ++j)
      for (int k = j + 1; k <= n; ++k) {
        int tmp1 = sum[i][j] - check(p[i], p[j], p[k]);
        int tmp2 = sum[j][k] - check(p[j], p[k], p[i]);
        int tmp3 = sum[i][k] - check(p[i], p[k], p[j]);
        int tot = Abs(tmp1 + tmp2 - tmp3);
        ans1 += tot * (n - 4);
        ans2 += tot * (tot - 1) / 2;
      }
  long long ans3 = (ans1 - ans2 * 4) / 2;
  long long ans =
      n * (long long)(n - 1) / 2 * (n - 2) / 3 * (n - 3) / 4 * (n - 4) / 5 -
      ans3 - ans2;
  printf("%lld", ans);
  return 0;
}
