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::prelude::*;

fn input<T: std::str::FromStr>() -> T {
    let stdin = std::io::stdin();
    let token: String = stdin
        .lock()
        .bytes()
        .map(|c| c.unwrap() as char)
        .skip_while(|c| c.is_whitespace())
        .take_while(|c| !c.is_whitespace())
        .collect();
    token.parse().ok().unwrap()
}

fn main() {
    let n: usize = input();
    let xy: Vec<(i32, i32)> = (0..n).map(|_| (input(), input())).collect();
    const V: usize = 100005;

    let mut e = vec![vec![vec![]; V]; 2];
    for (x, y) in xy {
        let (x, y) = (x as usize - 1, y as usize - 1);
        e[0][x].push(y);
        e[1][y].push(x);
    }

    let mut used = vec![vec![false; V]; 2];
    let mut ans = 0i64;
    for i in 0..V {
        if !used[0][i] {
            let mut cnt = vec![0, 0];
            dfs(i, 0, &e, &mut used, &mut cnt);
            ans += cnt[0] * cnt[1];
        }
        if !used[1][i] {
            let mut cnt = vec![0, 0];
            dfs(i, 0, &e, &mut used, &mut cnt);
            ans += cnt[0] * cnt[1];
        }
    }

    println!("{}", ans - n as i64);
}

fn dfs(
    cur: usize,
    xy: usize,
    e: &Vec<Vec<Vec<usize>>>,
    used: &mut Vec<Vec<bool>>,
    cnt: &mut Vec<i64>,
) {
    if used[xy][cur] {
        return;
    }

    used[xy][cur] = true;
    cnt[xy] += 1;
    for &next in &e[xy][cur] {
        let xy = 1 - xy;
        if !used[xy][next] {
            dfs(next, xy, e, used, cnt);
        }
    }
}