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 (Clang 3.8.0) (judge verdict: Wrong Answer)

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


int main(void)
{
	long int n;
	long int i, j;
	long int sum_x = 0;
	long int sum_y = 0;
	long int x[100000];
	long int y[100000];
	long int flg_x[100000] = {0};
	long int flg_y[100000] = {0};
	
	scanf("%ld", &n);
	
	for(i = 0; i < n; i++) {
		scanf("%ld%ld", &x[i], &y[i]);
		flg_x[x[i] - 1]++;
		flg_y[y[i] - 1]++;
	}
	
	if(n < 3) {
		printf("0\n");
		return 0;
	}
	
	for(i = 0; i < 100000; i++) {
		if(flg_x[i] != 0) {
			sum_x++;
		}
		if(flg_y[i] != 0) {
			sum_y++;
		}
	}
	
	if( (sum_x == 0) || (sum_y == 0) ) {
		printf("0\n");
		return 0;
	}
	
	printf("%ld\n", sum_x * sum_y - n);
	
	return 0;
}