#include<bits/stdc++.h>
#define reg register
typedef long long ll;
using namespace std;
const int MN=2e5+5;
int deg[MN][2];
int n,m;
char ch[MN];
int to[MN<<1],nxt[MN<<1],h[MN],cnt;
inline void ins(int s,int t){
	to[++cnt]=t;nxt[cnt]=h[s];h[s]=cnt;
	to[++cnt]=s;nxt[cnt]=h[t];h[t]=cnt;
}
int que[MN],ta;
int main(){
	scanf("%d%d",&n,&m);scanf("%s",ch+1);
	for(reg int i=1,s,t;i<=m;i++){
		scanf("%d%d",&s,&t);ins(s,t);
		deg[s][ch[t]-'A']++;deg[t][ch[s]-'A']++;
	}
	for(reg int i=1;i<=n;i++)
		if(!deg[i][0]||!deg[i][1])que[++ta]=i;
	for(reg int i=1;i<=ta;i++)
		for(reg int j=h[que[i]];j;j=nxt[j])
			if(deg[to[j]][0]&&deg[to[j]][1]&&!(--deg[to[j]][ch[que[i]]-'A']))
				que[++ta]=to[j];
	puts(ta<n?"Yes":"No");
	return 0;
}
### Prompt

Create a solution in Cpp for 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
#include<bits/stdc++.h>
#define reg register
typedef long long ll;
using namespace std;
const int MN=2e5+5;
int deg[MN][2];
int n,m;
char ch[MN];
int to[MN<<1],nxt[MN<<1],h[MN],cnt;
inline void ins(int s,int t){
	to[++cnt]=t;nxt[cnt]=h[s];h[s]=cnt;
	to[++cnt]=s;nxt[cnt]=h[t];h[t]=cnt;
}
int que[MN],ta;
int main(){
	scanf("%d%d",&n,&m);scanf("%s",ch+1);
	for(reg int i=1,s,t;i<=m;i++){
		scanf("%d%d",&s,&t);ins(s,t);
		deg[s][ch[t]-'A']++;deg[t][ch[s]-'A']++;
	}
	for(reg int i=1;i<=n;i++)
		if(!deg[i][0]||!deg[i][1])que[++ta]=i;
	for(reg int i=1;i<=ta;i++)
		for(reg int j=h[que[i]];j;j=nxt[j])
			if(deg[to[j]][0]&&deg[to[j]][1]&&!(--deg[to[j]][ch[que[i]]-'A']))
				que[++ta]=to[j];
	puts(ta<n?"Yes":"No");
	return 0;
}
```