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::collections::{BTreeMap, BTreeSet, VecDeque};

fn main() {
    let s = std::io::stdin();
    let mut sc = Scanner { stdin: s.lock() };
    let n: usize = sc.read();
    let mut x: Vec<u64> = vec![];
    let mut y: Vec<u64> = vec![];
    for _ in 0..n {
        x.push(sc.read());
        y.push(sc.read());
    }

    let sx = x
        .iter()
        .cloned()
        .collect::<BTreeSet<_>>()
        .into_iter()
        .enumerate()
        .map(|(i, v)| (v, i))
        .collect::<BTreeMap<_, _>>();
    let sy = y
        .iter()
        .cloned()
        .collect::<BTreeSet<_>>()
        .into_iter()
        .enumerate()
        .map(|(i, v)| (v, i))
        .collect::<BTreeMap<_, _>>();

    let mut graph = vec![vec![]; sx.len() + sy.len()];
    for i in 0..n {
        let x = *sx.get(&x[i]).unwrap();
        let y = *sy.get(&y[i]).unwrap() + sx.len();
        graph[x].push(y);
        graph[y].push(x);
    }

    let mut color = vec![0; sx.len() + sy.len()];
    let mut cur = 1;
    for t in 0..color.len() {
        if color[t] != 0 {
            continue;
        }

        let mut q = VecDeque::new();
        q.push_front(t);
        color[t] = cur;
        while let Some(v) = q.pop_front() {
            for &next in graph[v].iter() {
                if color[next] == 0 {
                    q.push_back(next);
                    color[next] = cur;
                }
            }
        }
        cur += 1;
    }

    let mut count = vec![(0, 0); cur];
    for (i, color) in color.into_iter().enumerate() {
        if i < sx.len() {
            count[color].0 += 1;
        } else {
            count[color].1 += 1;
        }
    }

    let mut all = 0;
    for (x, y) in count.into_iter() {
        all += x * y;
    }
    println!("{}", all - n);
}

pub struct Scanner<R> {
    stdin: R,
}

impl<R: std::io::Read> Scanner<R> {
    pub fn read<T: std::str::FromStr>(&mut self) -> T {
        use std::io::Read;
        let buf = self
            .stdin
            .by_ref()
            .bytes()
            .map(|b| b.unwrap())
            .skip_while(|&b| b == b' ' || b == b'\n' || b == b'\r')
            .take_while(|&b| b != b' ' && b != b'\n' && b != b'\r')
            .collect::<Vec<_>>();
        unsafe { std::str::from_utf8_unchecked(&buf) }
            .parse()
            .ok()
            .expect("Parse error.")
    }
    pub fn vec<T: std::str::FromStr>(&mut self, n: usize) -> Vec<T> {
        (0..n).map(|_| self.read()).collect()
    }
    pub fn chars(&mut self) -> Vec<char> {
        self.read::<String>().chars().collect()
    }
}