#include <bits/stdc++.h>
using namespace std;
int main() {
  long long n, i, j;
  cin >> n;
  vector<pair<double, double> > v(n);
  map<pair<double, double>, long long> mp;
  for (i = 0; i < n; i++) {
    double a, b;
    cin >> a >> b;
    mp[{a, b}] = i;
    v[i] = {a, b};
  }
  long long ans = 0;
  for (i = 0; i < n; i++) {
    for (j = i + 1; j < n; j++) {
      double a = (v[i].first + v[j].first) / 2.0;
      double b = (v[i].second + v[j].second) / 2.0;
      if (mp.find({a, b}) != mp.end()) {
        long long c = mp[{a, b}];
        if (c != i && c != j) ans++;
      }
    }
  }
  cout << ans;
}

### Prompt

Develop a solution in CPP 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() {
  long long n, i, j;
  cin >> n;
  vector<pair<double, double> > v(n);
  map<pair<double, double>, long long> mp;
  for (i = 0; i < n; i++) {
    double a, b;
    cin >> a >> b;
    mp[{a, b}] = i;
    v[i] = {a, b};
  }
  long long ans = 0;
  for (i = 0; i < n; i++) {
    for (j = i + 1; j < n; j++) {
      double a = (v[i].first + v[j].first) / 2.0;
      double b = (v[i].second + v[j].second) / 2.0;
      if (mp.find({a, b}) != mp.end()) {
        long long c = mp[{a, b}];
        if (c != i && c != j) ans++;
      }
    }
  }
  cout << ans;
}

```