Problem:
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Connected Components

Write a program which reads relations in a SNS (Social Network Service), and judges that given pairs of users are reachable each other through the network.

Input

In the first line, two integer $n$ and $m$ are given. $n$ is the number of users in the SNS and $m$ is the number of relations in the SNS. The users in the SNS are identified by IDs $0, 1, ..., n-1$.

In the following $m$ lines, the relations are given. Each relation is given by two integers $s$ and $t$ that represents $s$ and $t$ are friends (and reachable each other).

In the next line, the number of queries $q$ is given. In the following $q$ lines, $q$ queries are given respectively. Each query consists of two integers $s$ and $t$ separated by a space character.

Output

For each query, print "yes" if $t$ is reachable from $s$ through the social network, "no" otherwise.

Constraints

$2 \leq n \leq 100,000$

$0 \leq m \leq 100,000$

$1 \leq q \leq 10,000$

Sample Input

10 9
0 1
0 2
3 4
5 7
5 6
6 7
6 8
7 8
8 9
3
0 1
5 9
1 3

Sample Output

yes
yes
no

Solution in Scala (judge verdict: Time Limit Exceeded)

import scala.collection.mutable
import scala.io.StdIn._

case class Edge(from: Int, to: Int)

object Main extends App {
  val Array(n, m) = readLine().split(" ").map(_.toInt)
  val group: mutable.Seq[Int] = mutable.Seq.fill(n)(-1)
  var edges: Set[Edge] = Set()

  for (i <- 1 to m) {
    val Array(v1, v2) = readLine().split(" ").map(_.toInt)
    edges += Edge(v1, v2)
  }

  var id = 0
  while (edges.nonEmpty) {
    crawl(edges.head, id)
    id += 1
  }

  for (i <- 1 to readInt()) {
    val Array(v1, v2) = readLine().split(" ").map(_.toInt)
    if (group(v1) != -1 && group(v1) == group(v2)) println("yes") else println("no")
  }

  def crawl(edge: Edge, id: Int): Unit = {
    val que = mutable.Queue(edge)
    while (que.nonEmpty) {
      val e1 = que.dequeue()
      group(e1.to) = id
      group(e1.from) = id
      edges -= e1
      edges.foreach(e => {
        if (e.from == e1.from || e.to == e1.from || e.from == e1.to || e.to == e1.to)
          que.enqueue(e)
      })
    }
  }
}