#include <bits/stdc++.h>
using namespace std;
int main() {
  vector<pair<int, int>> a;
  int x, y;
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> x >> y;
    a.push_back(make_pair(x, y));
  }
  sort(a.begin(), a.end());
  int x1, y1, count = 0;
  for (int i = 0; i < n - 1; i++) {
    for (int j = i + 1; j < n; j++) {
      x1 = (a[i].first + a[j].first) / 2;
      y1 = (a[i].second + a[j].second) / 2;
      if (a[i].first + a[j].first != 2 * x1) continue;
      if (a[i].second + a[j].second != 2 * y1) continue;
      if (binary_search(a.begin(), a.end(), make_pair(x1, y1))) count++;
    }
  }
  cout << count;
  return 0;
}

### Prompt

Your challenge is to write a cpp solution to the following problem:

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() {
  vector<pair<int, int>> a;
  int x, y;
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> x >> y;
    a.push_back(make_pair(x, y));
  }
  sort(a.begin(), a.end());
  int x1, y1, count = 0;
  for (int i = 0; i < n - 1; i++) {
    for (int j = i + 1; j < n; j++) {
      x1 = (a[i].first + a[j].first) / 2;
      y1 = (a[i].second + a[j].second) / 2;
      if (a[i].first + a[j].first != 2 * x1) continue;
      if (a[i].second + a[j].second != 2 * y1) continue;
      if (binary_search(a.begin(), a.end(), make_pair(x1, y1))) count++;
    }
  }
  cout << count;
  return 0;
}

```