#include <bits/stdc++.h>
using namespace std;
int main() {
  int n;
  scanf("%d", &n);
  map<pair<int, int>, bool> mp;
  pair<int, int> v[n];
  for (int i = 0; i < n; i++) {
    int x, y;
    scanf("%d %d", &x, &y);
    mp[{x, y}] = true;
    v[i] = {x, y};
  }
  int cnt = 0;
  for (int i = 0; i < n; i++) {
    for (int j = i + 1; j < n; j++) {
      if ((v[i].first + v[j].first) % 2 or (v[i].second + v[j].second) % 2)
        continue;
      pair<int, int> mid = {(v[i].first + v[j].first) / 2,
                            (v[i].second + v[j].second) / 2};
      if (mp.count(mid)) cnt++;
    }
  }
  printf("%d\n", cnt);
  return 0;
}

### Prompt

Please formulate 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() {
  int n;
  scanf("%d", &n);
  map<pair<int, int>, bool> mp;
  pair<int, int> v[n];
  for (int i = 0; i < n; i++) {
    int x, y;
    scanf("%d %d", &x, &y);
    mp[{x, y}] = true;
    v[i] = {x, y};
  }
  int cnt = 0;
  for (int i = 0; i < n; i++) {
    for (int j = i + 1; j < n; j++) {
      if ((v[i].first + v[j].first) % 2 or (v[i].second + v[j].second) % 2)
        continue;
      pair<int, int> mid = {(v[i].first + v[j].first) / 2,
                            (v[i].second + v[j].second) / 2};
      if (mp.count(mid)) cnt++;
    }
  }
  printf("%d\n", cnt);
  return 0;
}

```