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)

#[allow(unused_imports)]
use std::cmp::*;
#[allow(unused_imports)]
use std::collections::*;
use std::io::{Write, BufWriter};
// https://qiita.com/tanakh/items/0ba42c7ca36cd29d0ac8
macro_rules! input {
    ($($r:tt)*) => {
        let stdin = std::io::stdin();
        let mut bytes = std::io::Read::bytes(std::io::BufReader::new(stdin.lock()));
        let mut next = move || -> String{
            bytes
                .by_ref()
                .map(|r|r.unwrap() as char)
                .skip_while(|c|c.is_whitespace())
                .take_while(|c|!c.is_whitespace())
                .collect()
        };
        input_inner!{next, $($r)*}
    };
}

macro_rules! input_inner {
    ($next:expr) => {};
    ($next:expr, ) => {};

    ($next:expr, $var:ident : $t:tt $($r:tt)*) => {
        let $var = read_value!($next, $t);
        input_inner!{$next $($r)*}
    };
}

macro_rules! read_value {
    ($next:expr, ( $($t:tt),* )) => {
        ( $(read_value!($next, $t)),* )
    };

    ($next:expr, [ $t:tt ; $len:expr ]) => {
        (0..$len).map(|_| read_value!($next, $t)).collect::<Vec<_>>()
    };

    ($next:expr, chars) => {
        read_value!($next, String).chars().collect::<Vec<char>>()
    };

    ($next:expr, usize1) => {
        read_value!($next, usize) - 1
    };

    ($next:expr, [ $t:tt ]) => {{
        let len = read_value!($next, usize);
        (0..len).map(|_| read_value!($next, $t)).collect::<Vec<_>>()
    }};

    ($next:expr, $t:ty) => {
        $next().parse::<$t>().expect("Parse error")
    };
}

/**
 * Union-Find tree.
 * Verified by https://atcoder.jp/contests/keyence2019/submissions/4071067
 */
struct UnionFind { disj: Vec<usize>, rank: Vec<usize> }

impl UnionFind {
    fn new(n: usize) -> Self {
        let disj = (0..n).collect();
        UnionFind { disj: disj, rank: vec![0; n] }
    }
    fn root(&mut self, x: usize) -> usize {
        if x != self.disj[x] {
            let par = self.disj[x];
            let r = self.root(par);
            self.disj[x] = r;
        }
        self.disj[x]
    }
    fn unite(&mut self, x: usize, y: usize) {
        let mut x = self.root(x);
        let mut y = self.root(y);
        if x == y { return }
        if self.rank[x] > self.rank[y] {
            std::mem::swap(&mut x, &mut y);
        }
        self.disj[x] = y;
        self.rank[y] = std::cmp::max(self.rank[y], self.rank[x] + 1);
    }
    #[allow(unused)]
    fn is_same_set(&mut self, x: usize, y: usize) -> bool {
        self.root(x) == self.root(y)
    }
}

fn solve() {
    let out = std::io::stdout();
    let mut out = BufWriter::new(out.lock());
    macro_rules! puts {
        ($($format:tt)*) => (write!(out,$($format)*).unwrap());
    }
    input! {
        n: usize,
        xy: [(usize, usize); n],
    }
    const W: usize = 100_100;
    let mut uf = UnionFind::new(2 * W);
    let mut row = vec![vec![]; W];
    let mut col = vec![vec![]; W];
    for &(x, y) in &xy {
        uf.unite(x, W + y);
        row[x].push(y);
        col[y].push(x);
    }
    for i in 0..W {
        if row[i].is_empty() { continue; }
        for j in 0..row[i].len() - 1 {
            let x = row[i][j];
            let y = row[i][j + 1];
            uf.unite(W + x, W + y);
        }
    }
    for i in 0..W {
        if col[i].is_empty() { continue; }
        for j in 0..col[i].len() - 1 {
            let x = col[i][j];
            let y = col[i][j + 1];
            uf.unite(x, y);
        }
    }
    let mut tot: i64 = 0;
    let mut sz = vec![(0, 0); 2 * W];
    for i in 0..2 * W {
        let r = uf.root(i);
        if i < W {
            sz[r].0 += 1;
        } else {
            sz[r].1 += 1;
        }
    }
    for i in 0..2 * W {
        let (x, y) = sz[i];
        tot += x * y;
    }
    puts!("{}\n", tot - n as i64);
}

fn main() {
    // In order to avoid potential stack overflow, spawn a new thread.
    let stack_size = 104_857_600; // 100 MB
    let thd = std::thread::Builder::new().stack_size(stack_size);
    thd.spawn(|| solve()).unwrap().join().unwrap();
}