#include <bits/stdc++.h>
int n, x[3000], y[3000], map[3001][3001] = {}, i, j, cnt = 0;
int main() {
  scanf("%d", &n);
  for (i = 0; i < n; i++) {
    scanf("%d%d", &x[i], &y[i]);
    x[i] += 1000, y[i] += 1000;
    map[x[i]][y[i]] = 1;
  }
  for (i = 0; i < n; i++)
    for (j = 0; j < n; j++)
      if (i != j && (x[i] + x[j]) % 2 == 0 && (y[i] + y[j]) % 2 == 0 &&
          map[(x[i] + x[j]) / 2][(y[i] + y[j]) / 2])
        cnt++;
  printf("%d\n", cnt / 2);
}

### 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>
int n, x[3000], y[3000], map[3001][3001] = {}, i, j, cnt = 0;
int main() {
  scanf("%d", &n);
  for (i = 0; i < n; i++) {
    scanf("%d%d", &x[i], &y[i]);
    x[i] += 1000, y[i] += 1000;
    map[x[i]][y[i]] = 1;
  }
  for (i = 0; i < n; i++)
    for (j = 0; j < n; j++)
      if (i != j && (x[i] + x[j]) % 2 == 0 && (y[i] + y[j]) % 2 == 0 &&
          map[(x[i] + x[j]) / 2][(y[i] + y[j]) / 2])
        cnt++;
  printf("%d\n", cnt / 2);
}

```