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)

//---------- begin union_find ----------
#[allow(dead_code)]
mod union_find {
    pub struct UF {
        p: Vec<i32>,
    }
    impl UF {
        pub fn new(n: usize) -> UF {
            UF {p: vec![-1; n] }
        }
        pub fn init(&mut self) {
            for p in self.p.iter_mut() {
                *p = -1;
            }
        }
        pub fn root(&self, mut x: usize) -> usize {
            while self.p[x] >= 0 {
                x = self.p[x] as usize;
            }
            x
        }
        pub fn same(&self, x: usize, y: usize) -> bool {
            self.root(x) == self.root(y)
        }
        pub fn unite(&mut self, x: usize, y: usize) -> Option<(usize, usize)> {
            let mut x = self.root(x);
            let mut y = self.root(y);
            if x == y {
                return None;
            }
            if self.p[x] > self.p[y] {
                std::mem::swap(&mut x, &mut y);
            }
            self.p[x] += self.p[y];
            self.p[y] = x as i32;
            Some((x, y))
        }
        pub fn parent(&self, x: usize) -> Option<usize> {
            let p = self.p[x];
            if p >= 0 {
                Some(p as usize)
            } else {
                None
            }
        }
        pub fn get_size(&self, x: usize) -> usize {
            let r = self.root(x);
            (-self.p[r]) as usize
        }
    }
}
//---------- end union_find ----------

use proconio::*;
use proconio::marker::*;

fn run() {
    input! {
        n: usize,
        p: [(Usize1, Usize1); n],
    }
    let w = 100_000;
    let mut dp = vec![(0usize, 1usize); 2 * w];
    for p in dp.iter_mut().take(w) {
        *p = (1, 0);
    }
    let mut u = union_find::UF::new(2 * w);
    for (x, y) in p {
        if let Some((a, b)) = u.unite(x, y + w) {
            dp[a].0 += dp[b].0;
            dp[a].1 += dp[b].1;
        }
    }
    let mut ans = 0;
    for (i, p) in dp.iter().enumerate() {
        if i == u.root(i) {
            ans += p.0 * p.1;
        }
    }
    ans -= n;
//    let ans = dp.into_iter().map(|p| p.0 * p.1).sum::<usize>() - n;
    println!("{}", ans);
}

fn main() {
    run();
}