#include<cstdio>
const int MaxN=200010,MaxM=200010;
int N,M;
char s[MaxN];
struct edge{int to;edge*next;}E[MaxM*2],*ne=E,*fir[MaxN];
void link(int u,int v){*ne=(edge){v,fir[u]};fir[u]=ne++;}
bool del[MaxN];
int fa[MaxN],fb[MaxN];
int stk[MaxN];
int main(){
	scanf("%d%d%s",&N,&M,s+1);
	for(int i=0,u,v;i<M;i++)scanf("%d%d",&u,&v),link(u,v),link(v,u);
	int*top=stk;
	for(int i=1;i<=N;i++){
		for(edge*e=fir[i];e;e=e->next)
			s[e->to]=='A'?fa[i]++:fb[i]++;
		if(!fa[i]||!fb[i])del[*top++=i]=1;
	}
	while(top>stk){
		int v=*--top;
		for(edge*e=fir[v];e;e=e->next)if(!del[e->to]){
			s[v]=='A'?fa[e->to]--:fb[e->to]--;
			if(!fa[e->to]||!fb[e->to])del[*top++=e->to]=1;
		}
	}
	for(int i=1;i<=N;i++)if(!del[i])return puts("Yes"),0;
	puts("No");
}
### Prompt

Develop a solution in CPP to the problem described below:

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<cstdio>
const int MaxN=200010,MaxM=200010;
int N,M;
char s[MaxN];
struct edge{int to;edge*next;}E[MaxM*2],*ne=E,*fir[MaxN];
void link(int u,int v){*ne=(edge){v,fir[u]};fir[u]=ne++;}
bool del[MaxN];
int fa[MaxN],fb[MaxN];
int stk[MaxN];
int main(){
	scanf("%d%d%s",&N,&M,s+1);
	for(int i=0,u,v;i<M;i++)scanf("%d%d",&u,&v),link(u,v),link(v,u);
	int*top=stk;
	for(int i=1;i<=N;i++){
		for(edge*e=fir[i];e;e=e->next)
			s[e->to]=='A'?fa[i]++:fb[i]++;
		if(!fa[i]||!fb[i])del[*top++=i]=1;
	}
	while(top>stk){
		int v=*--top;
		for(edge*e=fir[v];e;e=e->next)if(!del[e->to]){
			s[v]=='A'?fa[e->to]--:fb[e->to]--;
			if(!fa[e->to]||!fb[e->to])del[*top++=e->to]=1;
		}
	}
	for(int i=1;i<=N;i++)if(!del[i])return puts("Yes"),0;
	puts("No");
}
```