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: Accepted)

#include<bits/stdc++.h>
using namespace std;

using Graph=vector<vector<int>>;
int main(){
    int n;
    cin>>n;
    int m;
    cin>>m;
    string s;
    cin>>s;
    Graph g(n);
    for(int i=0;i<m;i++){
        int a,b;
        cin>>a>>b;
        a--,b--;
        g[a].push_back(b);
        g[b].push_back(a);
    }
    vector<int> acnt(n);
    vector<int> bcnt(n);
    vector<int> ng(n,false);
    stack<int> ngs;
    for(int i=0;i<n;i++){
        for(int j=0;j<g[i].size();j++){
            if(s[g[i][j]]=='A') acnt[i]++;
            if(s[g[i][j]]=='B') bcnt[i]++;
        }
        if(acnt[i]==0 || bcnt[i]==0){
            ng[i]=true;
            ngs.push(i);
        }
    }
    
    while(!ngs.empty()){
        int v=ngs.top();
        ngs.pop();
        for(int i=0;i<g[v].size();i++){
            if(s[v]=='A'){
                if((--acnt[g[v][i]])==0 && !ng[g[v][i]]){
                    ng[g[v][i]]=true;
                    ngs.push(g[v][i]);
                }
                
            }
            if(s[v]=='B'){
                if((--bcnt[g[v][i]])==0 && !ng[g[v][i]]){
                    ng[g[v][i]]=true;
                    ngs.push(g[v][i]);
                }
            }
        }
    }    
    //cerr<<"#"<<endl;
    //for(int i=0;i<n;i++) cerr<<ng[i]<<endl;
    cout<<(accumulate(ng.begin(),ng.end(),0)==n ? "No" : "Yes")<<endl;
    
    return 0;
}