Problem:
Score : 600 points

Problem Statement

There are N dots in a two-dimensional plane. The coordinates of the i-th dot are (x_i, y_i).

We will repeat the following operation as long as possible:

Choose four integers a, b, c, d (a \neq c, b \neq d) such that there are dots at exactly three of the positions (a, b), (a, d), (c, b) and (c, d), and add a dot at the remaining position.

We can prove that we can only do this operation a finite number of times. Find the maximum number of times we can do the operation.

Constraints

1 \leq N \leq 10^5

1 \leq x_i, y_i \leq 10^5

If i \neq j, x_i \neq x_j or y_i \neq y_j.

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N
x_1 y_1
:
x_N y_N

Output

Print the maximum number of times we can do the operation.

Sample Input 1

3
1 1
5 1
5 5

Sample Output 1

1

By choosing a = 1, b = 1, c = 5, d = 5, we can add a dot at (1, 5). We cannot do the operation any more, so the maximum number of operations is 1.

Sample Input 2

2
10 10
20 20

Sample Output 2

0

There are only two dots, so we cannot do the operation at all.

Sample Input 3

9
1 1
2 1
3 1
4 1
5 1
1 2
1 3
1 4
1 5

Sample Output 3

16

We can do the operation for all choices of the form a = 1, b = 1, c = i, d = j (2 \leq i,j \leq 5), and no more. Thus, the maximum number of operations is 16.

Solution in C (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int x[10001], y[100001];
int N;


int main() {
  int result;
  int i, j, k, cnt;
  scanf ("%d", &N);
  
  for (i=0;i<N;i++) {
      scanf("%d %d", &x[i], &y[i]);
  }
  
  result = 0;
  for (i=0;i<N;i++) {
    for (j=0;j<N;j++) {
      if (x[i]<x[j] && y[i]!=y[j]) {
        cnt=0;
        for (k=0;k<N;k++) {
          if (k!=i && k!=j) {
            if ((x[k]==x[i] && y[k]==y[j]) || (x[k]==x[j] && y[k]==y[i])) {
              cnt++;
              if (cnt>1) break;
            }
          }
        }
        if (cnt==1) result++;
      }
    }
  }
  
  
  
  printf("%d", result);
  return 0;
}