/*Lucky_Glass*/
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<vector>
using namespace std;
const int N=int(2e5);
vector<int> lnk[2*N+5];
int n,m;
char fla[N+5];
int tag[2*N+5];
void DFS(int u){
	tag[u]=1;
	for(int i=0;i<lnk[u].size();i++){
		int v=lnk[u][i];
		if(tag[v]==0)
			DFS(v);
		if(tag[v]==1){
			printf("Yes\n");
			exit(0);
		}
	}
	tag[u]=-1;
}
int main(){
	scanf("%d%d%s",&n,&m,fla+1);
	for(int i=0;i<m;i++){
		int u,v;scanf("%d%d",&u,&v);
		if(fla[u]==fla[v]){
			lnk[u].push_back(v+n);
			lnk[v].push_back(u+n);
		}
		else{
			lnk[u+n].push_back(v);
			lnk[v+n].push_back(u);
		}
	}
	for(int i=1;i<=n;i++)
		if(tag[i]==0)
			DFS(i);
	printf("No\n");
	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
/*Lucky_Glass*/
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<vector>
using namespace std;
const int N=int(2e5);
vector<int> lnk[2*N+5];
int n,m;
char fla[N+5];
int tag[2*N+5];
void DFS(int u){
	tag[u]=1;
	for(int i=0;i<lnk[u].size();i++){
		int v=lnk[u][i];
		if(tag[v]==0)
			DFS(v);
		if(tag[v]==1){
			printf("Yes\n");
			exit(0);
		}
	}
	tag[u]=-1;
}
int main(){
	scanf("%d%d%s",&n,&m,fla+1);
	for(int i=0;i<m;i++){
		int u,v;scanf("%d%d",&u,&v);
		if(fla[u]==fla[v]){
			lnk[u].push_back(v+n);
			lnk[v].push_back(u+n);
		}
		else{
			lnk[u+n].push_back(v);
			lnk[v+n].push_back(u);
		}
	}
	for(int i=1;i<=n;i++)
		if(tag[i]==0)
			DFS(i);
	printf("No\n");
	return 0;
}
```