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;
long long nCk(int n, int k) {
  long long ans = 1;
  for (int i = n; i >= n - k + 1; --i) ans *= i;
  for (int i = 1; i <= k; ++i) ans /= i;
  return ans;
}
const int MAXN = 305;
pair<long long, long long> pnt[MAXN];
int cnt[MAXN][MAXN];
int ccw(pair<long long, long long> p0, pair<long long, long long> p1,
        pair<long long, long long> p2) {
  long long cross = (p1.first - p0.first) * (p2.second - p0.second) -
                    (p2.first - p0.first) * (p1.second - p0.second);
  if (cross < 0) return -1;
  if (cross > 0) return +1;
  return 0;
}
bool lesss(pair<long long, long long> l, pair<long long, long long> r) {
  return (l.first != r.first ? l.first < r.first : l.second > r.second);
}
int get(int i, int j, int k) {
  if (lesss(pnt[j], pnt[i])) swap(i, j);
  if (lesss(pnt[k], pnt[j])) swap(j, k);
  if (lesss(pnt[j], pnt[i])) swap(i, j);
  if (ccw(pnt[i], pnt[k], pnt[j]) < 0)
    return cnt[i][k] - cnt[i][j] - cnt[j][k] - (pnt[j].first < pnt[k].first);
  else
    return cnt[i][j] + cnt[j][k] - cnt[i][k];
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  for (int i = 0; i <= n - 1; ++i) {
    cin >> pnt[i].first >> pnt[i].second;
  }
  sort(pnt, pnt + n);
  for (int i = 0; i <= n - 1; ++i)
    for (int j = i + 1; j <= n - 1; ++j) {
      for (int k = 0; k <= n - 1; ++k)
        if (k != i && k != j && pnt[i].first <= pnt[k].first &&
            pnt[k].first < pnt[j].first && ccw(pnt[i], pnt[j], pnt[k]) < 0)
          ++cnt[i][j];
    }
  long long total = 0;
  for (int i = 0; i <= n - 1; ++i)
    for (int j = i + 1; j <= n - 1; ++j)
      for (int k = j + 1; k <= n - 1; ++k) {
        long long tmp = get(i, j, k);
        total += tmp * (n - 3 - tmp);
      }
  total /= 2;
  total = nCk(n, 5) - total;
  cout << total << endl;
  return 0;
}
