#include <bits/stdc++.h>
using namespace std;
const int mx = 3e3 + 10, inf = 1e9 + 10;
int x[mx], y[mx];
bool book[mx][mx];
int main() {
  int n;
  long long ans = 0;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    scanf("%d%d", &x[i], &y[i]);
    x[i] += 1000, y[i] += 1000;
    book[x[i]][y[i]] = 1;
  }
  for (int i = 1; i < n; i++) {
    for (int j = i + 1; j <= n; j++) {
      int xx = x[i] + x[j], yy = y[i] + y[j];
      if (xx % 2 || yy % 2) continue;
      if (book[xx / 2][yy / 2]) ans++;
    }
  }
  printf("%lld\n", ans);
  return 0;
}

### Prompt

Please create a solution in cpp 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;
const int mx = 3e3 + 10, inf = 1e9 + 10;
int x[mx], y[mx];
bool book[mx][mx];
int main() {
  int n;
  long long ans = 0;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    scanf("%d%d", &x[i], &y[i]);
    x[i] += 1000, y[i] += 1000;
    book[x[i]][y[i]] = 1;
  }
  for (int i = 1; i < n; i++) {
    for (int j = i + 1; j <= n; j++) {
      int xx = x[i] + x[j], yy = y[i] + y[j];
      if (xx % 2 || yy % 2) continue;
      if (book[xx / 2][yy / 2]) ans++;
    }
  }
  printf("%lld\n", ans);
  return 0;
}

```