#include <bits/stdc++.h>
using namespace std;
int main() {
  cin.tie(0), ios_base::sync_with_stdio(false);
  int n;
  cin >> n;
  vector<pair<int, int>> v;
  for (int i = 0, x, y; i < n; i++) cin >> x >> y, v.push_back({x, y});
  sort(v.begin(), v.end());
  long long ans = 0;
  for (int i = 0; i < n; i++) {
    for (int j = i + 1; j < n; j++) {
      int x = v[i].first + v[j].first;
      int y = v[i].second + v[j].second;
      if (x % 2 or y % 2) continue;
      x = x / 2, y = y / 2;
      ans += upper_bound(v.begin(), v.end(), make_pair(x, y)) -
             lower_bound(v.begin(), v.end(), make_pair(x, y));
    }
  }
  cout << ans << endl;
}

### Prompt

Please provide a CPP coded solution to the problem described below:

You are given n points on a plane. All points are different.

Find the number of different groups of three points (A, B, C) such that point B is the middle of segment AC. 

The groups of three points are considered unordered, that is, if point B is the middle of segment AC, then groups (A, B, C) and (C, B, A) are considered the same.

Input

The first line contains a single integer n (3 ≤ n ≤ 3000) — the number of points. 

Next n lines contain the points. The i-th line contains coordinates of the i-th point: two space-separated integers xi, yi ( - 1000 ≤ xi, yi ≤ 1000).

It is guaranteed that all given points are different.

Output

Print the single number — the answer to the problem. 

Examples

Input

3
1 1
2 2
3 3


Output

1


Input

3
0 0
-1 0
0 1


Output

0

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
  cin.tie(0), ios_base::sync_with_stdio(false);
  int n;
  cin >> n;
  vector<pair<int, int>> v;
  for (int i = 0, x, y; i < n; i++) cin >> x >> y, v.push_back({x, y});
  sort(v.begin(), v.end());
  long long ans = 0;
  for (int i = 0; i < n; i++) {
    for (int j = i + 1; j < n; j++) {
      int x = v[i].first + v[j].first;
      int y = v[i].second + v[j].second;
      if (x % 2 or y % 2) continue;
      x = x / 2, y = y / 2;
      ans += upper_bound(v.begin(), v.end(), make_pair(x, y)) -
             lower_bound(v.begin(), v.end(), make_pair(x, y));
    }
  }
  cout << ans << endl;
}

```