#include <bits/stdc++.h>
using namespace std;
int main() {
  set<int> s[10001];
  int n;
  cin >> n;
  int x[n], y[n];
  for (int i = 0; i < n; i++) {
    cin >> x[i] >> y[i];
    if (x[i] < 0)
      s[x[i] + 3000].insert(y[i]);
    else
      s[x[i]].insert(y[i]);
  }
  long long ans = 0;
  for (int i = 0; i < n; i++)
    for (int j = i + 1; j < n; j++) {
      if (abs(x[i] - x[j]) % 2 == 0 && abs(y[i] - y[j]) % 2 == 0) {
        int t = abs(x[i] - x[j]) / 2 + min(x[i], x[j]);
        int z = abs(y[i] - y[j]) / 2 + min(y[i], y[j]);
        if (t < 0) {
          if (s[t + 3000].find(z) != s[t + 3000].end()) ans++;
        } else {
          if (s[t].find(z) != s[t].end()) ans++;
        }
      }
    }
  cout << ans;
}

### Prompt

Your task is to create 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() {
  set<int> s[10001];
  int n;
  cin >> n;
  int x[n], y[n];
  for (int i = 0; i < n; i++) {
    cin >> x[i] >> y[i];
    if (x[i] < 0)
      s[x[i] + 3000].insert(y[i]);
    else
      s[x[i]].insert(y[i]);
  }
  long long ans = 0;
  for (int i = 0; i < n; i++)
    for (int j = i + 1; j < n; j++) {
      if (abs(x[i] - x[j]) % 2 == 0 && abs(y[i] - y[j]) % 2 == 0) {
        int t = abs(x[i] - x[j]) / 2 + min(x[i], x[j]);
        int z = abs(y[i] - y[j]) / 2 + min(y[i], y[j]);
        if (t < 0) {
          if (s[t + 3000].find(z) != s[t + 3000].end()) ans++;
        } else {
          if (s[t].find(z) != s[t].end()) ans++;
        }
      }
    }
  cout << ans;
}

```