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)

#[allow(unused_imports)]
use {
    proconio::{fastout, input, marker::*},
    rand::Rng,
    std::cmp::*,
    std::collections::*,
    std::ops::*,
};

const N: usize = 100000;

fn main() {
    input! {
        n: usize,
        xy: [(Usize1, Usize1); n]
    }
    let mut to = vec![vec![]; 2 * N];
    for (x, y) in xy {
        to[x].push(N + y);
        to[N + y].push(x);
    }
    let mut visit = vec![false; 2 * N];
    let mut ans = 0;
    for i in 0..N {
        if visit[i] {
            continue;
        }
        let (x, y) = dfs(i, &to, &mut visit);
        ans += x * y;
    }
    ans -= n;
    println!("{}", ans);
}

fn dfs(v: usize, to: &Vec<Vec<usize>>, visit: &mut Vec<bool>) -> (usize, usize) {
    let mut x = 0;
    let mut y = 0;
    if v / N == 0 {
        x += 1;
    } else {
        y += 1;
    }
    visit[v] = true;
    for &u in to[v].iter() {
        if visit[u] {
            continue;
        }
        let (dx, dy) = dfs(u, to, visit);
        x += dx;
        y += dy;
    }
    (x, y)
}