#include <stdio.h>
#include <string.h>
#include <queue>
#define MN 200000
#define MM 200000*2
int n,m,cnt;
int d[MN+5][2];
int hd[MN+5],to[MM+5],nxt[MM+5];
char s[MN+5]; bool vis[MN+5];
std::queue<int> q;
void add(int u,int v){
	static int rn=0;
	to[rn]=v,nxt[rn]=hd[u],hd[u]=rn++;
	d[v][s[u]-'A']++;
}
int main(){
	memset(hd,0xff,sizeof(hd));
	scanf("%d%d%s",&n,&m,&s[1]); cnt=n;
	for(int i=1;i<=m;i++){
		static int u,v;
		scanf("%d%d",&u,&v);
		add(u,v);
		if(u!=v) add(v,u);
	}
	for(int i=1;i<=n;i++)
		if(d[i][0]==0||d[i][1]==0){
			q.push(i); vis[i]=true; cnt--;
		}
	while(!q.empty()){
		int u=q.front(); q.pop();
		for(int i=hd[u];~i;i=nxt[i]){
			if(vis[to[i]]) continue;
			if(--d[to[i]][s[u]-'A']==0){
				q.push(to[i]); vis[to[i]]=true; cnt--;
			}
		}
	}
	puts(cnt?"Yes":"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>
#include <string.h>
#include <queue>
#define MN 200000
#define MM 200000*2
int n,m,cnt;
int d[MN+5][2];
int hd[MN+5],to[MM+5],nxt[MM+5];
char s[MN+5]; bool vis[MN+5];
std::queue<int> q;
void add(int u,int v){
	static int rn=0;
	to[rn]=v,nxt[rn]=hd[u],hd[u]=rn++;
	d[v][s[u]-'A']++;
}
int main(){
	memset(hd,0xff,sizeof(hd));
	scanf("%d%d%s",&n,&m,&s[1]); cnt=n;
	for(int i=1;i<=m;i++){
		static int u,v;
		scanf("%d%d",&u,&v);
		add(u,v);
		if(u!=v) add(v,u);
	}
	for(int i=1;i<=n;i++)
		if(d[i][0]==0||d[i][1]==0){
			q.push(i); vis[i]=true; cnt--;
		}
	while(!q.empty()){
		int u=q.front(); q.pop();
		for(int i=hd[u];~i;i=nxt[i]){
			if(vis[to[i]]) continue;
			if(--d[to[i]][s[u]-'A']==0){
				q.push(to[i]); vis[to[i]]=true; cnt--;
			}
		}
	}
	puts(cnt?"Yes":"No");
}
```