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: Wrong Answer)

use std::collections::HashSet;
use std::fs;
use std::io::{stdin, Read};

macro_rules! parse {
    ($it: ident) => {
        $it.next().unwrap().parse().unwrap()
    };
    ($it: ident,2) => {
        (parse!($it), parse!($it))
    };
    ($it: ident,3) => {
        (parse!($it), parse!($it), parse!($it))
    };
    ($it: ident,4) => {
        (parse!($it), parse!($it), parse!($it), parse!($it))
    };
    ($it: ident, $n: expr) => {
        (&mut $it).take($n).map(|e| e.parse().unwrap()).collect();
    };
}

fn solve(s: &str) {
    let mut it = s.split_whitespace();
    let n: usize = parse!(it);
    let m: usize = 100010;
    let mut g = vec![vec![]; 2 * m];
    for _ in 0..n {
        let (x, y): (usize, usize) = parse!(it, 2);
        g[x - 1].push(y - 1 + m);
        g[y - 1 + m].push(x - 1);
    }
    let mut ret = 0;
    let mut seen: HashSet<usize> = HashSet::new();
    for i in 0..(2 * n) {
        if !seen.contains(&i) {
            let mut part = HashSet::new();
            dfs(&g, i, &mut part);
            ret += get_cap(&g, &part, m);
            seen.extend(part.iter());
        }
    }
    println!("{}", ret);
}

fn get_cap(g: &Vec<Vec<usize>>, vs: &HashSet<usize>, n: usize) -> usize {
    let nx = vs.iter().filter(|&&i| i < n).count();
    let ny = vs.len() - nx;
    let ne = vs.iter().map(|&i| g[i].len()).sum::<usize>() / 2;
    nx * ny - ne
}

fn dfs(g: &Vec<Vec<usize>>, u: usize, seen: &mut HashSet<usize>) {
    seen.insert(u);
    for &v in &g[u] {
        if !seen.contains(&v) {
            dfs(g, v, seen);
        }
    }
}

fn test() {
    let mut f = fs::File::open("foo.txt").unwrap();
    let mut s = String::new();
    f.read_to_string(&mut s).unwrap();
    solve(&s);
}

fn main() {
    let mut s = String::new();
    stdin().read_to_string(&mut s).unwrap();
    solve(&s);
}

#[cfg(test)]
mod tests {
    use super::*;
    #[test]
    fn test_main() {
        test();
    }
    #[test]
    fn test_input() {
        let s = "
        ";
        solve(s);
    }
}