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>
const int MAXN = 310;
struct vec {
  long long x, y;
  vec() { x = y = 0; }
  vec(long long a, long long b) { x = a, y = b; }
  long long operator^(vec b) const { return x * b.y - y * b.x; }
  long long operator*(vec b) const { return x * b.x + y * b.y; }
  long long norm2() const { return x * x + y * y; }
  long double norm() const { return sqrt(x * x + y * y); }
  vec operator-(vec b) const { return vec(x - b.x, y - b.y); }
  vec operator+(vec b) const { return vec(x + b.x, y + b.y); }
  vec operator*(long double k) const { return vec(x * k, y * k); }
  bool operator<(const vec& b) const { return (*this ^ b) > 0; }
  friend std::istream& operator>>(std::istream& in, vec& x) {
    in >> x.x >> x.y;
    return in;
  }
  friend std::ostream& operator<<(std::ostream& out, vec& x) {
    out << x.x << ' ' << x.y;
    return out;
  }
} ps[MAXN];
std::vector<std::tuple<vec, int, int> > A[4];
int n;
long long dp[MAXN][MAXN][5];
int get(vec x) {
  if (x.x >= 0 && x.y < 0) return 3;
  if (x.x < 0 && x.y <= 0) return 2;
  if (x.x >= 0 && x.y < 0) return 1;
  return 0;
}
int main() {
  std::ios_base::sync_with_stdio(false), std::cin.tie(0);
  std::cin >> n;
  for (int i = 1; i <= n; ++i) std::cin >> ps[i];
  for (int i = 1; i <= n; ++i)
    for (int j = 1; j <= n; ++j)
      if (i != j) {
        vec x = ps[j] - ps[i];
        A[get(x)].emplace_back(x, i, j);
      }
  for (int i = 0; i < 4; ++i) {
    std::sort(A[i].begin(), A[i].end());
    for (auto t : A[i]) {
      ++dp[std::get<1>(t)][std::get<2>(t)][0];
      for (int j = 1; j <= n; ++j)
        if (j != std::get<1>(t))
          for (int k = 0; k < 4; ++k)
            dp[j][std::get<2>(t)][k + 1] += dp[j][std::get<1>(t)][k];
    }
  }
  long long ans = 0;
  for (int i = 1; i <= n; ++i) ans += dp[i][i][4];
  std::cout << ans << std::endl;
  return 0;
}
