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;
struct P {
  int x, y;
  P() {}
  P(int _x, int _y) : x(_x), y(_y) {}
  P operator+(const P& a) const { return P(x + a.x, y + a.y); }
  P operator-(const P& a) const { return P(x - a.x, y - a.y); }
  long long dot(const P& a) const {
    return (long long)x * a.x + (long long)y * a.y;
  }
  long long crs(const P& a) const {
    return (long long)x * a.y - (long long)y * a.x;
  }
  P rot90() const { return P(-y, x); }
  bool operator<(const P& a) const { return x < a.x || x == a.x && y < a.y; }
  bool upper() const { return y > 0 || y == 0 && x > 0; }
  void get() {
    cin >> x;
    cin >> y;
  }
  void out() { printf("%d %d\n", x, y); }
};
int n;
P a[310];
bitset<310> b[310][310];
int s[310];
int main() {
  cin >> n;
  for (int i = 0; i < n; i++) a[i].get();
  for (int i = 0; i < n; i++)
    for (int j = 0; j < n; j++)
      for (int k = 0; k < n; k++)
        if ((a[j] - a[i]).crs(a[k] - a[i]) > 0) b[i][j][k] = 1;
  for (int i = 0; i < n; i++)
    for (int j = 0; j < n; j++)
      for (int k = 0; k < n; k++)
        if ((a[j] - a[i]).crs(a[k] - a[i]) > 0) {
          s[(b[i][j] & b[j][k] & b[k][i]).count()]++;
        }
  for (int i = 0; i < n; i++) s[i] /= 3;
  long long S = (long long)n * (n - 1) * (n - 2) * (n - 3) * (n - 4) / 120;
  long long T = 0;
  for (int i = 0; i < n; i++) T += (long long)s[i] * i * (n - 4);
  S -= T / 2;
  for (int i = 0; i < n; i++) S += (long long)s[i] * i * (i - 1) / 2;
  cout << S << endl;
  return 0;
}
