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 Rust (1.42.0) (judge verdict: Accepted)

const N: usize = 100000;

fn main() {
	let mut s = String::new();
	use std::io::Read;
	std::io::stdin().read_to_string(&mut s).unwrap();
	let mut s = s.split_whitespace();
	let n: usize = s.next().unwrap().parse().unwrap();
	let mut edges = vec![vec![]; 2 * N];
	for _ in 0..n {
		let x = s.next().unwrap().parse::<usize>().unwrap() - 1;
		let y = s.next().unwrap().parse::<usize>().unwrap() - 1 + N;
		edges[x].push(y);
		edges[y].push(x);
	}
	let mut visit = vec![false; 2 * N];
	let mut ans = 0;
	for i in 0..N {
		if visit[i] || edges[i].len() == 0 {
			continue;
		}
		let mut q = vec![i];
		let mut cnt = [0; 2];
		while let Some(v) = q.pop() {
			if visit[v] {
				continue;
			}
			visit[v] = true;
			cnt[v / N] += 1;
			for &to in edges[v].iter() {
				if !visit[to] {
					q.push(to);
				}
			}
		}
		ans += cnt[0] * cnt[1];
	}
	ans -= n;
	println!("{}", ans);
}