#include<stdio.h>
int h[200010],to[400010],nex[400010],M;
void add(int a,int b){
	M++;
	to[M]=b;
	nex[M]=h[a];
	h[a]=M;
}
char s[200010];
int c[200010],d[200010][2],st[200010];
bool del[200010],ins[200010];
int main(){
	int n,m,i,x,y;
	scanf("%d%d%s",&n,&m,s+1);
	for(i=1;i<=n;i++)c[i]=s[i]=='B';
	while(m--){
		scanf("%d%d",&x,&y);
		d[x][c[y]]++;
		add(x,y);
		if(x!=y){
			d[y][c[x]]++;
			add(y,x);
		}
	}
	M=0;
	for(i=1;i<=n;i++){
		if(d[i][0]==0||d[i][1]==0)ins[st[++M]=i]=1;
	}
	while(M){
		del[x=st[M--]]=1;
		ins[x]=0;
		for(i=h[x];i;i=nex[i]){
			if(!del[to[i]]){
				d[to[i]][c[x]]--;
				if((d[to[i]][0]==0||d[to[i]][1]==0)&&!ins[to[i]])st[++M]=to[i];
			}
		}
	}
	for(i=1;i<=n;i++){
		if(!del[i]){
			puts("Yes");
			return 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<stdio.h>
int h[200010],to[400010],nex[400010],M;
void add(int a,int b){
	M++;
	to[M]=b;
	nex[M]=h[a];
	h[a]=M;
}
char s[200010];
int c[200010],d[200010][2],st[200010];
bool del[200010],ins[200010];
int main(){
	int n,m,i,x,y;
	scanf("%d%d%s",&n,&m,s+1);
	for(i=1;i<=n;i++)c[i]=s[i]=='B';
	while(m--){
		scanf("%d%d",&x,&y);
		d[x][c[y]]++;
		add(x,y);
		if(x!=y){
			d[y][c[x]]++;
			add(y,x);
		}
	}
	M=0;
	for(i=1;i<=n;i++){
		if(d[i][0]==0||d[i][1]==0)ins[st[++M]=i]=1;
	}
	while(M){
		del[x=st[M--]]=1;
		ins[x]=0;
		for(i=h[x];i;i=nex[i]){
			if(!del[to[i]]){
				d[to[i]][c[x]]--;
				if((d[to[i]][0]==0||d[to[i]][1]==0)&&!ins[to[i]])st[++M]=to[i];
			}
		}
	}
	for(i=1;i<=n;i++){
		if(!del[i]){
			puts("Yes");
			return 0;
		}
	}
	puts("No");
}
```