#include<cstdio>
int n,m,d[200050],p[200050][2],head[200050],q[200050],t,num=1,vis[200050];
struct r{
	int to,last;
}e[400050];
void add(int u,int v){
	e[num].to=v;e[num].last=head[u];head[u]=num++;
}
int main(){
	scanf("%d%d",&n,&m);
	for (int i=1;i<=n;i++){
		int c=getchar();
		while (c<'A'||c>'B')c=getchar();
		d[i]=c-'A';
	}
	int u,v;
	for (int i=1;i<=m;i++){
		scanf("%d%d",&u,&v);
		p[u][d[v]]++;p[v][d[u]]++;
		add(u,v);add(v,u);
	}
	for (int i=1;i<=n;i++)if (!p[i][0]||!p[i][1])q[++t]=i,vis[i]=1;
	for (int h=1;h<=t;h++){
		int x=q[h];
		for (int i=head[x];i;i=e[i].last)if (!vis[e[i].to]){	
			p[e[i].to][d[x]]--;
			if (!p[e[i].to][d[x]])q[++t]=e[i].to,vis[e[i].to]=1;
		}
	}
	if (t<n)printf("Yes");
	else printf("No");
}
### Prompt

Please formulate 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
#include<cstdio>
int n,m,d[200050],p[200050][2],head[200050],q[200050],t,num=1,vis[200050];
struct r{
	int to,last;
}e[400050];
void add(int u,int v){
	e[num].to=v;e[num].last=head[u];head[u]=num++;
}
int main(){
	scanf("%d%d",&n,&m);
	for (int i=1;i<=n;i++){
		int c=getchar();
		while (c<'A'||c>'B')c=getchar();
		d[i]=c-'A';
	}
	int u,v;
	for (int i=1;i<=m;i++){
		scanf("%d%d",&u,&v);
		p[u][d[v]]++;p[v][d[u]]++;
		add(u,v);add(v,u);
	}
	for (int i=1;i<=n;i++)if (!p[i][0]||!p[i][1])q[++t]=i,vis[i]=1;
	for (int h=1;h<=t;h++){
		int x=q[h];
		for (int i=head[x];i;i=e[i].last)if (!vis[e[i].to]){	
			p[e[i].to][d[x]]--;
			if (!p[e[i].to][d[x]])q[++t]=e[i].to,vis[e[i].to]=1;
		}
	}
	if (t<n)printf("Yes");
	else printf("No");
}
```