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: Accepted)

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

typedef struct {
	size_t n, c;
	int *arr;
} list;

void add(list *l, int x) {
	if (l->n+1 > l->c) {
		l->c = l->c*2 + !l->c;
		l->arr = realloc(l->arr, l->c * sizeof *l->arr);
	}
	l->arr[l->n++] = x;
}

int seenx[100000], seeny[100000];
list lstx[100000], lsty[100000];
long long int cntx, cnty;

void dfs(int i, long long int* cnt, long long int* cntb, int* seen, int* seenb, list *lst, list *lstb) {
	seen[i] = 1;
	(*cnt)++;
	for (size_t j = 0; j < lst[i].n; j++) {
		int k = lst[i].arr[j];
		if (seenb[k]) continue;
		dfs(k, cntb, cnt, seenb, seen, lstb, lst);
	}
}

int main(void) {
	int n;
	scanf("%d", &n);

	for (int i = 0; i < n; i++) {
		int x, y;
		scanf("%d%d", &x, &y);
		add(&lstx[x-1], y-1);
		add(&lsty[y-1], x-1);
	}

	long long int cnt = -n;
	for (int i = 0; i < 100000; i++) {
		if (seenx[i] || lstx[i].n == 0) continue;
		cntx = cnty = 0;
		dfs(i, &cntx, &cnty, seenx, seeny, lstx, lsty);
		cnt += cntx * cnty;
	}

	for (int i = 0; i < 100000; i++) {
		free(lstx[i].arr);
		free(lsty[i].arr);
	}

	printf("%lld\n", cnt);
}