#include <bits/stdc++.h>
using namespace std;
int const N = 3000 + 20, MAX = 2000 + 20;
int n, x[N], y[N], cnt[MAX][MAX];
long long ans;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 0; i < n; i++)
    cin >> x[i] >> y[i], cnt[x[i] + 1000][y[i] + 1000]++;
  for (int i = 0; i < n; i++)
    for (int j = i + 1; j < n; j++) {
      if (((x[i] + x[j]) & 1) || ((y[i] + y[j]) & 1)) continue;
      ans += cnt[(x[i] + x[j]) / 2 + 1000][(y[i] + y[j]) / 2 + 1000];
    }
  return cout << ans, 0;
}

### Prompt

Create a solution in CPP for 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 const N = 3000 + 20, MAX = 2000 + 20;
int n, x[N], y[N], cnt[MAX][MAX];
long long ans;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 0; i < n; i++)
    cin >> x[i] >> y[i], cnt[x[i] + 1000][y[i] + 1000]++;
  for (int i = 0; i < n; i++)
    for (int j = i + 1; j < n; j++) {
      if (((x[i] + x[j]) & 1) || ((y[i] + y[j]) & 1)) continue;
      ans += cnt[(x[i] + x[j]) / 2 + 1000][(y[i] + y[j]) / 2 + 1000];
    }
  return cout << ans, 0;
}

```