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: Accepted)


class DisjointSet 

public
	def initialize(size = nil)
		@rank = Array.new(size){0}
		@p = Array.new(size){0}
		if size 
			size.times do |i|
				@p[i] = i
			end
		end
	end

	def same(x, y)
		return findSet(x) == findSet(y)
	end

	def unite(x, y)
		link(findSet(x), findSet(y))
	end

private
	def link(x, y)
		if @rank[x] > @rank[y]
			@p[y] = x
		else
			@p[x] = y
			@rank[y] += 1 if @rank[x] == @rank[y]
		end
	end

	def findSet(x)
		@p[x] = findSet(@p[x]) unless x == @p[x] 
		return @p[x]
	end

end

 
n = gets.split.map(&:to_i)
tree = DisjointSet.new(n[0])
n[1].times do
	a = gets.split.map(&:to_i)
	tree.unite(a[0], a[1])
end

n = gets.to_i

n.times do
	a = gets.split.map(&:to_i)
	if tree.same(a[0], a[1])
		puts "yes"
	else
		puts "no"
	end
end