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 C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<iostream>
#include<iomanip>
//#include<cstdio>
#include<vector>
#include<map>
#include<queue>
#include<algorithm>
#include<cmath>
#include<cassert>
#include<set>
using namespace std;
typedef long long ll;
const int Nmax=2e5;
int N, M, A[Nmax], B[Nmax];
string S;
vector<vector<int> > graph;
vector<int> state;

int dfs(int n, int s){
	if(state[n] >= 0) return 0;
	state[n] = s;
	for(int m: graph[n]){
		//cout << n << ":" << s << ":" << r << "->";
		//cout << m << endl;
		if(s==3 && state[m]==0) return 1;
		if((s==0&&S[m]=='A')||
			(s==1&&S[m]=='B')||
			(s==2&&S[m]=='B')||
			(s==3&&S[m]=='A')){
			if(dfs(m, (s+1)%4)) return 1;
		}
	}
	return 0;
}

int main(){
	cin >> N >> M >> S;
	for(int i=0; i<M; i++){
		cin >> A[i] >> B[i];
		A[i]--; B[i]--;
	}
	S = S+S;

	graph = vector<vector<int> >(2*N);
	for(int i=0; i<M; i++){
		if(A[i]!=B[i]){
			graph[A[i]].push_back(B[i]);
			graph[B[i]].push_back(A[i]);
			graph[A[i]+N].push_back(B[i]);
			graph[B[i]+N].push_back(A[i]);
		} else {
			graph[A[i]].push_back(A[i]+N);
		}
	}

	state = vector<int>(2*N, -1);
	for(int n=0; n<N; n++){
		if(S[n]=='A' && dfs(n, 0)){
			cout << "Yes" << endl;
			return 0;
		}
	}

	cout << "No" << endl;
	return 0;
}