#include <bits/stdc++.h>
using namespace std;
const int MOD = 1e9 + 7;
bool ok[2001][2001];
int main() {
  {
    ios_base::sync_with_stdio(false);
    cin.tie(0);
  }
  int n;
  cin >> n;
  memset(ok, false, sizeof(ok));
  pair<int, int> a[n];
  for (int i = 0; i <= n - 1; i++) {
    cin >> a[i].first >> a[i].second;
    ok[a[i].first + 1000][a[i].second + 1000] = true;
  }
  int res = 0;
  for (int i = 0; i <= n - 1; i++) {
    for (int j = i + 1; j <= n - 1; j++) {
      int x = a[i].first + a[j].first;
      int y = a[i].second + a[j].second;
      if (x & 1 || y & 1) continue;
      if (ok[x / 2 + 1000][y / 2 + 1000]) res++;
    }
  }
  cout << res;
  return 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;
const int MOD = 1e9 + 7;
bool ok[2001][2001];
int main() {
  {
    ios_base::sync_with_stdio(false);
    cin.tie(0);
  }
  int n;
  cin >> n;
  memset(ok, false, sizeof(ok));
  pair<int, int> a[n];
  for (int i = 0; i <= n - 1; i++) {
    cin >> a[i].first >> a[i].second;
    ok[a[i].first + 1000][a[i].second + 1000] = true;
  }
  int res = 0;
  for (int i = 0; i <= n - 1; i++) {
    for (int j = i + 1; j <= n - 1; j++) {
      int x = a[i].first + a[j].first;
      int y = a[i].second + a[j].second;
      if (x & 1 || y & 1) continue;
      if (ok[x / 2 + 1000][y / 2 + 1000]) res++;
    }
  }
  cout << res;
  return 0;
}

```