#include<bits/stdc++.h>
using namespace std;
#define REP(i,a,b) for(int i=(a),ed=(b);i<=ed;++i)
typedef long long ll;
inline int read(){
    register int x=0,f=1;register char ch=getchar();
    while(!isdigit(ch)){if(ch=='-')f=0;ch=getchar();}
    while(isdigit(ch)){x=x*10+(ch^'0');ch=getchar();}
    return f?x:-x; 
}

const int N=2e5+10;
int n,m,cnt[N][2],del[N];
vector<int> E[N];
char s[N];

inline void Dfs(int u){
    REP(i,0,E[u].size()-1){
	int v=E[u][i];
	--cnt[v][s[u]-'A'];
	if(!del[v]&&!cnt[v][s[u]-'A'])del[v]=1,Dfs(v);
    }
}

int main(){
    //freopen("in.in","r",stdin);
    n=read(),m=read();scanf("%s",s+1);
    REP(i,1,m){int u=read(),v=read();E[u].push_back(v),E[v].push_back(u);++cnt[u][s[v]-'A'],++cnt[v][s[u]-'A'];}
    REP(i,1,n)if(!del[i]&&!(cnt[i][0]&&cnt[i][1]))del[i]=1,Dfs(i);
    REP(i,1,n)if(!del[i]){puts("Yes");return 0;}
    puts("No");
    return 0;
}

### Prompt

Your task is to create a CPP solution to the following problem:

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`.

Examples

Input

2 3
AB
1 1
1 2
2 2


Output

Yes


Input

4 3
ABAB
1 2
2 3
3 1


Output

No


Input

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


Output

Yes


Input

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


Output

No

### Response

```cpp
#include<bits/stdc++.h>
using namespace std;
#define REP(i,a,b) for(int i=(a),ed=(b);i<=ed;++i)
typedef long long ll;
inline int read(){
    register int x=0,f=1;register char ch=getchar();
    while(!isdigit(ch)){if(ch=='-')f=0;ch=getchar();}
    while(isdigit(ch)){x=x*10+(ch^'0');ch=getchar();}
    return f?x:-x; 
}

const int N=2e5+10;
int n,m,cnt[N][2],del[N];
vector<int> E[N];
char s[N];

inline void Dfs(int u){
    REP(i,0,E[u].size()-1){
	int v=E[u][i];
	--cnt[v][s[u]-'A'];
	if(!del[v]&&!cnt[v][s[u]-'A'])del[v]=1,Dfs(v);
    }
}

int main(){
    //freopen("in.in","r",stdin);
    n=read(),m=read();scanf("%s",s+1);
    REP(i,1,m){int u=read(),v=read();E[u].push_back(v),E[v].push_back(u);++cnt[u][s[v]-'A'],++cnt[v][s[u]-'A'];}
    REP(i,1,n)if(!del[i]&&!(cnt[i][0]&&cnt[i][1]))del[i]=1,Dfs(i);
    REP(i,1,n)if(!del[i]){puts("Yes");return 0;}
    puts("No");
    return 0;
}

```