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 Scala (2.11.7) (judge verdict: Wrong Answer)

import java.io._
import java.util.StringTokenizer
import scala.util.control.Breaks.{breakable,break}
import scala.collection.mutable.Queue

object Main extends App {
  def comb2(x: Long): Long = {
    if(x <= 1)
      0
    else
      (x * (x-1)) / 2
  }

  val in = new InputReader(System.in)
  val pw = new PrintWriter(System.out)

  val max = 1e5.toInt

  val N = in.next().toInt
  val X = Array.fill[List[Int]](max+1)(Nil)
  val Y = Array.fill[List[Int]](max+1)(Nil)
  for (_ <- 1 to N) {
    val x, y = in.next().toInt
    X(x) = y :: X(x)
    Y(y) = x :: Y(y)
  }

  //
  val dpX = new Array[Boolean](max+1)
  val dpY = new Array[Boolean](max+1)

  var ans = 0L
  for(i <- 1 to N) {
    if(!dpX(i)) {
      var nodex: List[Int] = Nil
      var nodey: List[Int] = Nil
      val q = new Queue[(Int, String)]

      dpX(i) = true
      q.enqueue((i, "x"))
      nodex = i :: nodex

      while(!q.isEmpty) {
        val (n, label) = q.dequeue()

        if(label == "x") {
          for(e <- X(n) if !dpY(e)) {
            dpY(e) = true
            q.enqueue((e, "y"))
            nodey = e :: nodey
          }
        }
        else {
          for(e <- Y(n) if !dpX(e)) {
            dpX(e) = true
            q.enqueue((e, "x"))
            nodex = e :: nodex
          }
        }
      }

      val cntx = nodex.length
      val cnty = nodey.length
      if (cntx >= 2 && cnty >= 2) {
        var edges = 0

        for (e <- nodex) {
          edges += X(e).length
        }

        ans += cntx * cnty - edges
      }
    }
  }

  println(ans)
}


class InputReader(stream: InputStream) {
  private val reader = new BufferedReader(new InputStreamReader(stream))
  private var tokenizer: StringTokenizer = new StringTokenizer(reader.readLine())
  def next(): String = {
    while (!tokenizer.hasMoreTokens()) {
      tokenizer = new StringTokenizer(reader.readLine())
    }
    tokenizer.nextToken()
  }
}