Problem:
Disjoint Set

Write a program which manipulates a disjoint set S = {S1, S2, . . . , Sk}.

First of all, the program should read an integer n, then make a disjoint set where each element consists of 0, 1, ... n−1 respectively.

Next, the program should read an integer q and manipulate the set for q queries. There are two kinds of queries for different operations:

unite(x, y): unites sets that contain x and y, say Sx and Sy, into a new set.

same(x, y): determine whether x and y are in the same set.

Input

n q
com1 x1 y1
com2 x2 y2
...
comq xq yq

In the first line, n and q are given. Then, q queries are given where com represents the type of queries. '0' denotes unite and '1' denotes same operation.

Output

For each same operation, print 1 if x and y are in the same set, otherwise 0, in a line.

Constraints

1 ≤ n ≤ 10000

1 ≤ q ≤ 100000

x ≠ y

Sample Input

5 12
0 1 4
0 2 3
1 1 2
1 3 4
1 1 4
1 3 2
0 1 3
1 2 4
1 3 0
0 0 4
1 0 2
1 3 0

Sample Output

0
0
1
1
1
0
1
1

Solution in Scala (judge verdict: Accepted)

import io.StdIn

object Main extends App {

  val out = new java.io.PrintWriter(System.out)
  val Array(n, q) = StdIn.readLine.split(" ").map(_.toInt)
  val djs = new DisjointSet(n)

  (0 until q).foreach { _ =>
    val Array(com, x, y) = StdIn.readLine.split(" ").map(_.toInt)
    if(com == 0) djs.unite(x, y)
    else out.println(if(djs.same(x, y)) "1" else "0")
  }

  out.flush

  class DisjointSet(size: Int) {
    var rank, p = Array.fill[Int](size)(0)

    (0 until size).foreach(i => makeSet(i))

    def makeSet(x: Int): Unit = {
      p(x) = x
      rank(x) = 0
    }
    def findSet(x: Int): Int = {
      if(x != p(x)) p(x) = findSet(p(x))
      p(x)
    }
    def unite(x: Int, y: Int): Unit = {
      def link(x: Int, y: Int): Unit = {
        if(rank(x) > rank(x)) p(y) = x
        else {
          p(x) = y
          if(rank(x) == rank(y)) rank(y) += 1
        }
      }

      link(findSet(x), findSet(y))
    }
    def same(x: Int, y: Int): Boolean = {
      findSet(x) == findSet(y)
    }
  }
}