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: Time Limit Exceeded)

# unionfindぽいけどもっと単純に答えるらしい
# って思ってやったんだけどTLEなるじゃん は？？？？？？？？？？
# なんかリストを作るときに探索して仲間わけしてるからクエリの判定がO(1)で済むっていうやり方をやってるらしい
n, m = gets.chomp.split.map(&:to_i)
arr = Array.new(n).map{[]}
group = Array.new(n, 0) # これで同じグラフに属するか判断するらしい
# ただし判断するときに、0同士は独立した者同士ってことを考慮する
id = 1  # グループ番号
m.times do
  a, b = gets.chomp.split.map(&:to_i)
  arr[a] << b
  arr[b] << a
  # ここで探索して仲間わけするらしい
  stack = [a]
  if group[a].zero?
    group[a] = id
  end
  until stack.empty?
    nex = stack.pop
    arr[nex].each do |i|
      next if group[i] == id
      group[i] = id
      stack << i
    end
  end
  id+=1
end
q = gets.chomp.to_i
q.times do
  a, b = gets.chomp.split.map(&:to_i)
  if group[a] == group[b] && group[a] > 0 && group[b] > 0 # 0同士は独立した者同士
    puts "yes"
  else
    puts "no"
  end
end