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;

// Input macro from https://qiita.com/tanakh/items/0ba42c7ca36cd29d0ac8
macro_rules! input {
  (source = $s:expr, $($r:tt)*) => {
    let mut iter = $s.split_whitespace();
    input_inner!{iter, $($r)*}
  };
  ($($r:tt)*) => {
    #[allow(unused_mut)]
    let mut s = {
      use std::io::Read;
      let mut s = String::new();
      std::io::stdin().read_to_string(&mut s).unwrap();
      s
    };
    let mut iter = s.split_whitespace();
    input_inner!{iter, $($r)*}
  };
}

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

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

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

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

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

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

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

const K: usize = 100_000;

fn dfs(i: usize, edge: &Vec<Vec<usize>>, used: &mut Vec<bool>, count: &mut [i64; 2]) -> i64 {
  used[i] = true;
  count[i / K] += 1;
  for &j in &edge[i] {
    if ! used[j] {
      dfs(j, edge, used, count);
    }
  }
  count[0] * count[1]
}

fn main() {
  input! {
    n: usize,
    p: [(usize1, usize1); n],
  }
  let mut edge = vec![vec![]; 2 * K];
  for &(x, y) in &p {
    let y = y + K;
    edge[x].push(y);
    edge[y].push(x);
  }
  let mut used = vec![false; 2 * K];
  let mut ans = 0;
  for i in 0 .. 2 * K {
    ans += dfs(i, &edge, &mut used, &mut [0, 0]);
  }

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