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 Ruby (judge verdict: Wrong Answer)

class DisjointSet
  attr_reader :parent, :rank

  def initialize(ids = [])
    @parent, @rank = {}, {}
    ids.each do |id|
      @parent[id] = id
      @rank[id] = 0
    end
  end

  def make_set(x)
    @parent[x] = x
    @rank[x] = 0
  end

  def find(x)
    return x if (@parent[x] == x)
    @parent[x] = find(@parent[x])
    @parent[x]
  end

  def union(x, y)
    root_x, root_y, rank_x, rank_y =
      find(x), find(y), @rank[x], @rank[y]

    return if (root_x == root_y)

    if (rank_x > rank_y)
      @parent[y] = root_x
    elsif (rank_x < rank_y)
      @parent[x] = root_y
    else
      @parent[y] = root_x
      @rank[x] += 1
    end
  end

  def same?(x, y)
    find(x) == find(y)
  end
end

n, m = gets.chop.split.map(&:to_i)

g = DisjointSet.new

n.times do |i|
  g.make_set i
end

m.times do
  g.union *gets.chop.split.map(&:to_i)
end

q = gets.chop.to_i

q.times do
  puts (g.same?(*gets.chop.split.map(&:to_i))) ? 'yes' : 'no'
end