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)

fn read_stdin() -> String {
    use std::io::Read;
    let mut input = String::new();
    std::io::stdin().read_to_string(&mut input).unwrap();
    input
}

fn scan<'a, I: Iterator<Item = &'a str>, T: std::str::FromStr>(tokens: &mut I) -> T {
    tokens.next().unwrap().parse().ok().unwrap()
}

fn main() {
    let input = read_stdin();
    let tokens = &mut input.split_whitespace();
    let n: usize = scan(tokens);
    let mut solver = Solver::new(100001, n);
    for _ in 0..n {
        solver.add_point(scan(tokens), scan(tokens));
    }
    solver.solve();
}

struct Solver {
    n: usize,
    k: usize,
    ys_of_x: Vec<Vec<usize>>,
    xs_of_y: Vec<Vec<usize>>,
    c_of_x: Vec<usize>,
    c_of_y: Vec<usize>,
}

impl Solver {
    fn new(n: usize, k: usize) -> Solver {
        Solver {
            n: n,
            k: k,
            ys_of_x: vec![Vec::new(); n],
            xs_of_y: vec![Vec::new(); n],
            c_of_x: vec![0; n],
            c_of_y: vec![0; n],
        }
    }
    fn add_point(&mut self, x: usize, y: usize) {
        self.ys_of_x[x].push(y);
        self.xs_of_y[y].push(x);
    }
    fn solve(&mut self) {
        let mut m = 0;
        for x in 0..self.n {
            if self.c_of_x[x] == 0 && !self.ys_of_x[x].is_empty() {
                m += 1;
                self.dfs_x(x, m);
            }
        }
        let mut nx_of_c: Vec<usize> = vec![0; m + 1];
        let mut ny_of_c: Vec<usize> = vec![0; m + 1];
        for x in 0..self.n {
            nx_of_c[self.c_of_x[x]] += 1;
        }
        for y in 0..self.n {
            ny_of_c[self.c_of_y[y]] += 1;
        }
        let mut ans = 0;
        for c in 1..m + 1 {
            ans += nx_of_c[c] * ny_of_c[c];
        }
        ans -= self.k;
        println!("{}", ans);
    }
    fn dfs_x(&mut self, x: usize, c: usize) {
        if self.c_of_x[x] != 0 {
            return;
        }
        self.c_of_x[x] = c;
        for i in 0..self.ys_of_x[x].len() {
            let y = self.ys_of_x[x][i];
            self.dfs_y(y, c);
        }
    }
    fn dfs_y(&mut self, y: usize, c: usize) {
        if self.c_of_y[y] != 0 {
            return;
        }
        self.c_of_y[y] = c;
        for i in 0..self.xs_of_y[y].len() {
            let x = self.xs_of_y[y][i];
            self.dfs_x(x, c);
        }
    }
}