#include<bits/stdc++.h>
#define rg register  
const int MN=400000+5;
using namespace std;
typedef long long ll;
template<typename T>inline T&IN(T&in){
	in=0;char c=getchar();int f=1;
	while(!isdigit(c)){if(c=='-')f=-1;c=getchar();}
	while(isdigit(c))in=in*10+c-'0',c=getchar();
	return in*=f;
}
queue<int>q;
int n,ans,m;char c[MN];
int d[MN][2],vis[MN],w[MN];
int nxt[MN],to[MN],h[MN],cnt;
void add(int u,int v){to[cnt]=v,nxt[cnt]=h[u],h[u]=cnt++;}
void input(){
	IN(n);IN(m);memset(h,-1,sizeof h);ans=n;
	scanf("%s",c+1);
	for(int i=1;i<=n;++i)w[i]=c[i]-'A';
	for(int i=1,u,v;i<=m;++i){
		IN(u),IN(v);add(u,v),add(v,u);
		d[v][w[u]]++,d[u][w[v]]++;
	}
	for(int i=1;i<=n;++i)if(!d[i][0]||!d[i][1])vis[i]=1,q.push(i),ans--;
	while(!q.empty()){
		int u=q.front();q.pop();
		for(int e=h[u],v=to[e];~e;e=nxt[e],v=to[e]){
			d[v][w[u]]--;
			if(!d[v][w[u]]&&!vis[v])q.push(v),ans--,vis[v]=1;
		}
	}
	if(ans)puts("Yes");
	else puts("No");
}
int main(){
	input();
	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>
#define rg register  
const int MN=400000+5;
using namespace std;
typedef long long ll;
template<typename T>inline T&IN(T&in){
	in=0;char c=getchar();int f=1;
	while(!isdigit(c)){if(c=='-')f=-1;c=getchar();}
	while(isdigit(c))in=in*10+c-'0',c=getchar();
	return in*=f;
}
queue<int>q;
int n,ans,m;char c[MN];
int d[MN][2],vis[MN],w[MN];
int nxt[MN],to[MN],h[MN],cnt;
void add(int u,int v){to[cnt]=v,nxt[cnt]=h[u],h[u]=cnt++;}
void input(){
	IN(n);IN(m);memset(h,-1,sizeof h);ans=n;
	scanf("%s",c+1);
	for(int i=1;i<=n;++i)w[i]=c[i]-'A';
	for(int i=1,u,v;i<=m;++i){
		IN(u),IN(v);add(u,v),add(v,u);
		d[v][w[u]]++,d[u][w[v]]++;
	}
	for(int i=1;i<=n;++i)if(!d[i][0]||!d[i][1])vis[i]=1,q.push(i),ans--;
	while(!q.empty()){
		int u=q.front();q.pop();
		for(int e=h[u],v=to[e];~e;e=nxt[e],v=to[e]){
			d[v][w[u]]--;
			if(!d[v][w[u]]&&!vis[v])q.push(v),ans--,vis[v]=1;
		}
	}
	if(ans)puts("Yes");
	else puts("No");
}
int main(){
	input();
	return 0;
}
```