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;
using ll = long long;
using ii = pair<int, int>;
using vi = vector<int>;
struct pt {
  ll x, y;
  pt operator-(const pt &o) const { return {x - o.x, y - o.y}; }
};
ll cross(pt &a, pt &b) { return a.x * b.y - b.x * a.y; }
int side(pt &p) {
  assert(p.x or p.y);
  return p.y < 0 or p.y == 0 and p.y < 0;
}
bool cmp(pt &p, pt &q) {
  return side(p) == side(q) ? cross(p, q) > 0 : side(p) < side(q);
}
const int N = 302;
ll dp[6][N][N];
int main(int argc, char const *argv[]) {
  cin.sync_with_stdio(0);
  cin.tie(0);
  cin.exceptions(cin.failbit);
  int n;
  cin >> n;
  vector<pt> v(n);
  for (pt &p : v) {
    cin >> p.x >> p.y;
  }
  vector<ii> es;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++)
      if (i != j) {
        es.push_back({i, j});
      }
  }
  sort(es.begin(), es.end(), [&](ii a, ii b) {
    pt z = {0, 0};
    pt p = v[a.second] - v[a.first];
    pt q = v[b.second] - v[b.first];
    return cmp(p, q);
  });
  for (int i = 0; i < n; i++) {
    dp[0][i][i] = 1;
  }
  for (ii &e : es) {
    int s = e.first, t = e.second;
    for (int k = 0; k < 5; k++) {
      for (int i = 0; i < n; i++) {
        dp[k + 1][i][t] += dp[k][i][s];
      }
    }
  }
  ll ans = 0;
  for (int i = 0; i < n; i++) {
    ans += dp[5][i][i];
  }
  cout << ans << endl;
  return 0;
}
