#include <bits/stdc++.h>
using namespace std;
int n, x[3005], y[3005];
set<pair<int, int> > S;
int main() {
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> x[i] >> y[i];
    S.insert(make_pair(x[i] * 2, y[i] * 2));
  }
  int ans = 0;
  for (int i = 0; i < n; i++)
    for (int j = i + 1; j < n; j++)
      if (S.count(make_pair(x[i] + x[j], y[i] + y[j]))) ans++;
  cout << ans << endl;
  return 0;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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 n, x[3005], y[3005];
set<pair<int, int> > S;
int main() {
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> x[i] >> y[i];
    S.insert(make_pair(x[i] * 2, y[i] * 2));
  }
  int ans = 0;
  for (int i = 0; i < n; i++)
    for (int j = i + 1; j < n; j++)
      if (S.count(make_pair(x[i] + x[j], y[i] + y[j]))) ans++;
  cout << ans << endl;
  return 0;
}

```