#include <bits/stdc++.h>
using namespace std;
int main() {
  int ans = 0, n;
  cin >> n;
  vector<int> x(n, 0), y(n, 0);
  vector<vector<int>> m(2001, vector<int>(2001));
  for (int i = 0; i < n; i++) {
    cin >> x[i] >> y[i];
    m[x[i] += 1000][y[i] += 1000] = 1;
  }
  for (int i = 0; i < n; i++) {
    for (int j = i + 1; j < n; j++) {
      int x1 = (x[i] + x[j]), y1 = (y[i] + y[j]);
      if (x1 % 2 == 0 && y1 % 2 == 0 && m[x1 / 2][y1 / 2] == 1) ans++;
    }
  }
  cout << ans;
}

### 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() {
  int ans = 0, n;
  cin >> n;
  vector<int> x(n, 0), y(n, 0);
  vector<vector<int>> m(2001, vector<int>(2001));
  for (int i = 0; i < n; i++) {
    cin >> x[i] >> y[i];
    m[x[i] += 1000][y[i] += 1000] = 1;
  }
  for (int i = 0; i < n; i++) {
    for (int j = i + 1; j < n; j++) {
      int x1 = (x[i] + x[j]), y1 = (y[i] + y[j]);
      if (x1 % 2 == 0 && y1 % 2 == 0 && m[x1 / 2][y1 / 2] == 1) ans++;
    }
  }
  cout << ans;
}

```