Problem:
Score : 900 points

Problem Statement

You are given an undirected graph consisting of N vertices and M edges.
The vertices are numbered 1 to N, and the edges are numbered 1 to M.
In addition, each vertex has a label, A or B. The label of Vertex i is s_i.
Edge i bidirectionally connects vertex a_i and b_i.

The phantom thief Nusook likes to choose some vertex as the startpoint and traverse an edge zero or more times.
Today, he will make a string after traveling as above, by placing the labels of the visited vertices in the order visited, beginning from the startpoint.

For example, in a graph where Vertex 1 has the label A and Vertex 2 has the label B, if Nusook travels along the path 1 \rightarrow 2 \rightarrow 1 \rightarrow 2 \rightarrow 2, the resulting string is ABABB.

Determine if Nusook can make all strings consisting of A and B.

Constraints

2 \leq N \leq 2 \times 10^{5}

1 \leq M \leq 2 \times 10^{5}

|s| = N

s_i is A or B.

1 \leq a_i, b_i \leq N

The given graph may NOT be simple or connected.

Input

Input is given from Standard Input in the following format:

N M
s
a_1 b_1
:
a_{M} b_{M}

Output

If Nusook can make all strings consisting of A and B, print Yes; otherwise, print No.

Sample Input 1

2 3
AB
1 1
1 2
2 2

Sample Output 1

Yes

Since Nusook can visit Vertex 1 and Vertex 2 in any way he likes, he can make all strings consisting of A and B.

Sample Input 2

4 3
ABAB
1 2
2 3
3 1

Sample Output 2

No

For example, Nusook cannot make BB.

The given graph may not be connected.

Sample Input 3

13 23
ABAAAABBBBAAB
7 1
10 6
1 11
2 10
2 8
2 11
11 12
8 3
7 12
11 2
13 13
11 9
4 1
9 7
9 6
8 13
8 6
4 10
8 7
4 3
2 1
8 12
6 9

Sample Output 3

Yes

Sample Input 4

13 17
BBABBBAABABBA
7 1
7 9
11 12
3 9
11 9
2 1
11 5
12 11
10 8
1 11
1 8
7 7
9 10
8 8
8 12
6 2
13 11

Sample Output 4

No

Solution in Ruby (2.3.3) (judge verdict: Accepted)

def is_ok(ev,val,v,srchd,ok)
	srchd[v] = true
	flga = false
	flgb = false
	ev.each{|v_of_ev|
		if(ok[v_of_ev])
			case val[v_of_ev]
			when "A"
				flga = true
			when "B"
				flgb = true
			end
		end
		if(flga && flgb)
			break
		end
	}
	if(!flga || !flgb)
		# break this vertex
		ok[v] = false
		return false
	end
	return true
end
def search(e,val)
	stack = [] # stack
	srchd = Array.new(e.size){false} # is searched flag
	ok = Array.new(e.size){true}
	# init
	e.each_with_index{|ev,v|
		if(!srchd[v] && !is_ok(ev,val,v,srchd,ok))
			chain(e,val,v,srchd,ok)
		end
	} 
	# detect cyclic
	# passed = Array.new(e.size){false} # is searched flag
	# for i in 0..(N-1)
	# 	if(!passed[i])
	# 		stack = [] # stack
	# 		e[i].each{|v|
	# 			stack.push([v,s]) #v,d to v
	# 		} 
	# 		while(!stack.empty?)
	# 			curv,curp = stack.pop
	# 			if(passed[curv])
	# 				# return true
	# 			end
	# 			passed[curv]=true
	# 			e[curv].each{|vv|
	# 				stack.push([vv,v])
	# 			}
	# 		end
	# 	end
	# end
	for i in 0..(N-1)
		return true if ok[i]
	end
	return false
end

def chain(e,val, s,srchd,ok)
	stack = [] # stack
	# init
	e[s].each{|v|
		stack.push([v,s]) #v,d to v
	} 
	while(!stack.empty?)
		curv,curp = stack.pop
		srchd[curv]=true
		if(!is_ok(e[curv],val,curv,srchd,ok))
			e[curv].each{|vv|
				if(vv != curp && ok[vv])
					stack.push([vv,curv])
				end
			} 
		end
	end
end


N,M=gets.chomp.split(' ').map{|n| n.to_i}
Val=gets.chomp.split('')
E = Array.new(N){[]}
1.upto(M){
	p=gets.chomp.split(' ').map{|n| n.to_i}
	p[0]-=1
	p[1]-=1
	E[p[0]].push(p[1])
	E[p[1]].push(p[0])
}
if(search(E,Val))
	puts "Yes"
else
	puts "No"
end