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 9.2.1) (judge verdict: Wrong Answer)

#include <stdio.h>

int main()
{
	int i, N, x[100001], y[100001];
	scanf("%d", &N);
	for (i = 1; i <= N; i++) scanf("%d %d", &(x[i]), &(y[i]));

	int n[2] = {}, count[2][100001] = {};
	for (i = 1; i <= N; i++) {
		if (count[0][x[i]] == 0) n[0]++;
		if (count[1][y[i]] == 0) n[1]++;
		count[0][x[i]]++;
		count[1][y[i]]++;
	}
	
	long long ans = (long long)n[0] * n[1] - N;
	for (i = 1; i <= N; i++) {
		if (count[0][x[i]] == 1 && count[1][y[i]] == 1) {
			ans -= n[0] + n[1] - 2;
			n[0]--;
			n[1]--;
		}
	}	
	
	printf("%lld\n", ans);
	fflush(stdout);
	return 0;
}