#include<bits/stdc++.h>
using namespace std;
const int N=800005,M=998244353;
int n,m,u,v,i,head[N],Next[N*2],adj[N*2],d[N],q[N],l,r,k;
char c[200005];
void Push(int u,int v)
{
	Next[++k]=head[u];
	head[u]=k;
	adj[k]=v;
	++d[v];
}
int main()
{
	scanf("%d %d",&n,&m);
	scanf("%s",c+1);
	for(i=1;i<=m;++i)
	{
		scanf("%d %d",&u,&v);
		if(c[u]=='A'&&c[v]=='A')
			Push(u,v+n);
		if(c[u]=='A'&&c[v]=='B')
			Push(u+n,v+2*n);
		if(c[u]=='B'&&c[v]=='B')
			Push(u+2*n,v+3*n);
		if(c[u]=='B'&&c[v]=='A')
			Push(u+3*n,v);
		swap(u,v);
		if(c[u]=='A'&&c[v]=='A')
			Push(u,v+n);
		if(c[u]=='A'&&c[v]=='B')
			Push(u+n,v+2*n);
		if(c[u]=='B'&&c[v]=='B')
			Push(u+2*n,v+3*n);
		if(c[u]=='B'&&c[v]=='A')
			Push(u+3*n,v);
	}
	for(i=1;i<=n*4;++i)
		if(!d[i])
			q[++r]=i;
	l=1;
	while(l<=r)
	{
		for(i=head[q[l]];i!=0;i=Next[i])
		{
			--d[adj[i]];
			if(d[adj[i]]==0)
				q[++r]=adj[i];
		}
		++l;
	}
	puts(r==4*n?"No":"Yes");
}

### Prompt

Develop a solution in CPP to the problem described below:

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>
using namespace std;
const int N=800005,M=998244353;
int n,m,u,v,i,head[N],Next[N*2],adj[N*2],d[N],q[N],l,r,k;
char c[200005];
void Push(int u,int v)
{
	Next[++k]=head[u];
	head[u]=k;
	adj[k]=v;
	++d[v];
}
int main()
{
	scanf("%d %d",&n,&m);
	scanf("%s",c+1);
	for(i=1;i<=m;++i)
	{
		scanf("%d %d",&u,&v);
		if(c[u]=='A'&&c[v]=='A')
			Push(u,v+n);
		if(c[u]=='A'&&c[v]=='B')
			Push(u+n,v+2*n);
		if(c[u]=='B'&&c[v]=='B')
			Push(u+2*n,v+3*n);
		if(c[u]=='B'&&c[v]=='A')
			Push(u+3*n,v);
		swap(u,v);
		if(c[u]=='A'&&c[v]=='A')
			Push(u,v+n);
		if(c[u]=='A'&&c[v]=='B')
			Push(u+n,v+2*n);
		if(c[u]=='B'&&c[v]=='B')
			Push(u+2*n,v+3*n);
		if(c[u]=='B'&&c[v]=='A')
			Push(u+3*n,v);
	}
	for(i=1;i<=n*4;++i)
		if(!d[i])
			q[++r]=i;
	l=1;
	while(l<=r)
	{
		for(i=head[q[l]];i!=0;i=Next[i])
		{
			--d[adj[i]];
			if(d[adj[i]]==0)
				q[++r]=adj[i];
		}
		++l;
	}
	puts(r==4*n?"No":"Yes");
}

```