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 OCaml (4.02.3) (judge verdict: Accepted)

let n = read_int ();;

type unionfind = { size : int array; parent : int array };;
let init_unionfind n = {size = Array.make n 1; parent = Array.init n (fun i -> i)};;
let rec find uf x = (
    if uf.parent.(x) = x then
      x
    else (
      let y = find uf (uf.parent.(x)) in
      uf.parent.(x) <- y;
      y
    )
  );;
let unite uf x y = (
    let x = find uf x in
    let y = find uf y in
    if x <> y then (
      let x, y = if uf.size.(x) > uf.size.(y) then x, y else y, x in
      uf.size.(x) <- uf.size.(x) + uf.size.(y);
      uf.parent.(y) <- x;
    );
  );;

let uf = init_unionfind 200000;;

for i = 0 to n - 1 do
  let x, y = Scanf.scanf " %d %d" (fun x y -> x - 1, y - 1 + 100000) in
  unite uf x y
done

let row = Array.make 200000 0;;
let col = Array.make 200000 0;;

for i = 0 to 99999 do
  let j = find uf i in
  row.(j) <- row.(j) + 1;
done;;

for i = 100000 to 199999 do
  let j = find uf i in
  col.(j) <- col.(j) + 1;
done;;

let ans = ref (-n);;
for i = 0 to 199999 do
  ans := !ans + row.(i) * col.(i);
done;;

Printf.printf "%d\n" !ans;;