#include<bits/stdc++.h>
#define neko 400010
#define f(i,a,b) for(int i=(a);i<=(b);++i)
using namespace std;
int book[neko],fa[neko],t,head[neko],vis[neko],n,m,ans;
char s[neko];
struct node
{
	int v,w,nex;
}e[neko<<1];
void add(int x,int y,int z)
{
	e[++t].v=y,e[t].w=z,e[t].nex=head[x],head[x]=t;
	e[++t].v=x,e[t].w=z,e[t].nex=head[y],head[y]=t;
}
void dfs(int u,int col)
{
	int tag=0;
	//printf("%d %d\n",u,col);
	for(register int i=head[u],v=e[i].v;i;i=e[i].nex,v=e[i].v)
	{
		if(!book[i]&&e[i].w^col)
		{
			book[i]=1,vis[i]=1;
			dfs(v,e[i].w);
			book[i]=0;
		}
		else if(book[i]&&e[i].w^col)
		{
			printf("Yes\n");exit(0);
		}
	}
}
int main()
{
	int x,y;
	scanf("%d%d",&n,&m);
	scanf("%s",s+1);
	f(i,1,m)
	{
		scanf("%d%d",&x,&y);
		if(s[x]^s[y])add(x,y,0);
		else add(x,y,1);
	}
	f(i,1,m<<1)if(!vis[i])vis[i]=1,dfs(e[i].v,e[i].w);
	printf("No\n");
	return 0;
}
### 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<bits/stdc++.h>
#define neko 400010
#define f(i,a,b) for(int i=(a);i<=(b);++i)
using namespace std;
int book[neko],fa[neko],t,head[neko],vis[neko],n,m,ans;
char s[neko];
struct node
{
	int v,w,nex;
}e[neko<<1];
void add(int x,int y,int z)
{
	e[++t].v=y,e[t].w=z,e[t].nex=head[x],head[x]=t;
	e[++t].v=x,e[t].w=z,e[t].nex=head[y],head[y]=t;
}
void dfs(int u,int col)
{
	int tag=0;
	//printf("%d %d\n",u,col);
	for(register int i=head[u],v=e[i].v;i;i=e[i].nex,v=e[i].v)
	{
		if(!book[i]&&e[i].w^col)
		{
			book[i]=1,vis[i]=1;
			dfs(v,e[i].w);
			book[i]=0;
		}
		else if(book[i]&&e[i].w^col)
		{
			printf("Yes\n");exit(0);
		}
	}
}
int main()
{
	int x,y;
	scanf("%d%d",&n,&m);
	scanf("%s",s+1);
	f(i,1,m)
	{
		scanf("%d%d",&x,&y);
		if(s[x]^s[y])add(x,y,0);
		else add(x,y,1);
	}
	f(i,1,m<<1)if(!vis[i])vis[i]=1,dfs(e[i].v,e[i].w);
	printf("No\n");
	return 0;
}
```