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)

//Problem B
#include<bits/stdc++.h>
using namespace std;
const int N=400005;
int n,m,cnt,tot,dfn[N],low[N],st[N],tp,in[N];
char ch[N];
vector<int>e[N],E[N];
void tarjan(int v){
    dfn[v]=low[v]=++cnt;
    st[++tp]=v,in[v]=1;
    for(int i=0;i<(int)E[v].size();i++){
        int u=E[v][i];
        if(!dfn[u])tarjan(u),low[v]=min(low[v],low[u]);
        else if(in[u])low[v]=min(low[v],dfn[u]);
    }
    if(dfn[v]==low[v]){
        ++tot;int lust=0;
        while(lust!=v)lust=st[tp--],in[lust]=0;
    }
}
int main(){
    scanf("%d%d%s",&n,&m,ch+1);
    while(m--){
        int a,b;
        scanf("%d%d",&a,&b);
        e[a].push_back(b),e[b].push_back(a);
        if(ch[a]==ch[b])E[a].push_back(b+n),E[b].push_back(a+n);
        else E[a+n].push_back(b),E[b+n].push_back(a);
    }
    n<<=1;
    for(int i=1;i<=n;i++)if(!dfn[i])tarjan(i);
    if(tot==n)printf("No");else printf("Yes");
    //for(int i=1;i<=n;i++)for(int j=0;j<2;j++)if(!vis[i][j]){printf("try %d %d\n",i,j);
    //    ++cnt;if(dfs(i,j)){printf("Yes");for(int i=1;i<=n;i++){for(int j=0;j<2;j++)printf("%d ",vis[i][j]);printf("\n");}return 0;}
    //}printf("No");
    return 0;
}