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;
const int N = 301;
long long f[N][N][5];
int n;
struct point {
  int x, y;
  point(int X = 0, int Y = 0) { x = X, y = Y; }
  int id() {
    if (x >= 0) {
      if (y >= 0)
        return 0;
      else
        return 3;
    } else if (y >= 0)
      return 1;
    else
      return 2;
  }
  long long operator*(point a) {
    return (long long)this->x * a.y - (long long)this->y * a.x;
  }
  bool operator<(point a) {
    if (this->id() != a.id())
      return this->id() < a.id();
    else
      return *this * a >= 0;
  }
  point operator-(point a) { return point(this->x - a.x, this->y - a.y); }
} a[N];
vector<pair<int, int> > d;
bool cmp(pair<int, int> x, pair<int, int> y) {
  return (a[x.second] - a[x.first]) < (a[y.second] - a[y.first]);
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d %d", &a[i].x, &a[i].y);
    f[i][i][0] = 1;
  }
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= n; j++)
      if (i != j) d.push_back(make_pair(i, j));
  sort(d.begin(), d.end(), cmp);
  for (auto u : d)
    for (int i = 0; i < 4; i++)
      for (int j = 1; j <= n; j++)
        if (f[j][u.first][i] && j != u.second &&
            (a[u.second] - a[u.first]) < (a[j] - a[u.second])) {
          f[j][u.second][i + 1] += f[j][u.first][i];
        }
  long long ans = 0;
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= n; j++) ans += f[i][j][4];
  cout << ans;
  return 0;
}
