#include<cstdio>
#include<cstring>
#include<iostream>
#include<queue>
using namespace std;
const int MAXX=200100;
int head[MAXX],ver[MAXX<<1],nxt[MAXX<<1],ind[MAXX][2];
int n,m,tot,cnt;
bool vis[MAXX];
char s[MAXX];
queue<int>q;
inline void add(int x,int y){
	s[x]=='A' ? ++ind[y][0] : ++ind[y][1];
	ver[++tot]=y;
	nxt[tot]=head[x];
	head[x]=tot;
}
inline void topsort(){
	for(int i=1;i<=n;++i)if((!ind[i][0])||(!ind[i][1])){
		q.push(i);
		vis[i]=1;
		cnt++;
	}
	while(!q.empty()){
		int x=q.front();
		q.pop();
		for(int i=head[x];i;i=nxt[i]){
			int y=ver[i];
			s[x]=='A' ? --ind[y][0] : --ind[y][1];
            if((!ind[y][0])||(!ind[y][1])){
            	if(vis[y])continue;
            	vis[y]=1;
            	q.push(y);
            	cnt++;
            }
		}
	}
}
int main(){
	scanf("%d%d",&n,&m);
	scanf("%s",s+1);
    for(int i=1;i<=m;++i){
    	int x,y;
    	scanf("%d%d",&x,&y);
    	add(x,y);
    	add(y,x);
    }
    topsort();
    if(cnt==n)printf("No\n");
    else printf("Yes\n");
    return 0;
}
### Prompt

Construct a CPP code solution to the problem outlined:

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>
#include<cstring>
#include<iostream>
#include<queue>
using namespace std;
const int MAXX=200100;
int head[MAXX],ver[MAXX<<1],nxt[MAXX<<1],ind[MAXX][2];
int n,m,tot,cnt;
bool vis[MAXX];
char s[MAXX];
queue<int>q;
inline void add(int x,int y){
	s[x]=='A' ? ++ind[y][0] : ++ind[y][1];
	ver[++tot]=y;
	nxt[tot]=head[x];
	head[x]=tot;
}
inline void topsort(){
	for(int i=1;i<=n;++i)if((!ind[i][0])||(!ind[i][1])){
		q.push(i);
		vis[i]=1;
		cnt++;
	}
	while(!q.empty()){
		int x=q.front();
		q.pop();
		for(int i=head[x];i;i=nxt[i]){
			int y=ver[i];
			s[x]=='A' ? --ind[y][0] : --ind[y][1];
            if((!ind[y][0])||(!ind[y][1])){
            	if(vis[y])continue;
            	vis[y]=1;
            	q.push(y);
            	cnt++;
            }
		}
	}
}
int main(){
	scanf("%d%d",&n,&m);
	scanf("%s",s+1);
    for(int i=1;i<=m;++i){
    	int x,y;
    	scanf("%d%d",&x,&y);
    	add(x,y);
    	add(y,x);
    }
    topsort();
    if(cnt==n)printf("No\n");
    else printf("Yes\n");
    return 0;
}
```