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 (1.15.1) (judge verdict: Accepted)

use std::io::*;

mod disjoint_set {
    pub struct DisjointSet {
        parent: Vec<usize>,
        size: Vec<usize>,
    }
    impl DisjointSet {
        pub fn new(n: usize) -> DisjointSet {
            DisjointSet {
                parent: (0..n).collect(),
                size: vec![1; n],
            }
        }
        pub fn root(&mut self, x: usize) -> usize {
            if x == self.parent[x] {
                x
            } else {
                let tmp = self.parent[x];
                self.parent[x] = self.root(tmp);
                self.parent[x]
            }
        }
        pub fn merge(&mut self, x: usize, y: usize) {
            let x = self.root(x);
            let y = self.root(y);
            if x == y {
                return;
            }
            if self.size[x] > self.size[y] {
                self.size[x] += self.size[y];
                self.parent[y] = x;
            } else {
                self.size[y] += self.size[x];
                self.parent[x] = y;
            }
        }
    }
}

fn main() {
    let mut s: String = String::new();
    std::io::stdin().read_to_string(&mut s).ok();
    let mut itr = s.trim().split_whitespace();
    let n: usize = itr.next().unwrap().parse().unwrap();
    let a: Vec<(usize, usize)> = (0..n)
        .map(|_| {
            (
                itr.next().unwrap().parse::<usize>().unwrap() - 1,
                itr.next().unwrap().parse::<usize>().unwrap() - 1,
            )
        })
        .collect();

    let mut ds = disjoint_set::DisjointSet::new(200020);
    let mut count = vec![0; 200020];
    let mut ans = 0;
    for i in 0..n {
        ds.merge(a[i].0, a[i].1 + 100000);
    }
    for i in 0..100000 {
        count[ds.root(i)] += 1;
    }
    for i in 100000..200000 {
        ans += count[ds.root(i)];
    }
    println!("{}", ans - n);
}

// この解法ではすでに長方形ができてる場合を余分に数えてる。
// すでにできてるかの判定が面倒なので、ありうる辺の数を数えて最後にnを引きたい。
// これは二部グラフの各連結成分を完全グラフにした時の辺の数。連結成分内なら絶対に長方形が作れる。
// 逆に連結成分でない頂点と長方形を作ろうとすると新たに辺が２本必要になるので無理。