#include<iostream>
#include<cstring>
#include<cstdio>
#include<queue>
#define next sfhds

using namespace std;

int a[200001][2],n,m,to[500001],next[500001],first[500001],is[200001],e1=0,k=0,v[200001];
char ch[200001];
queue<int> q;

void insert(int o,int p) {to[e1]=p;next[e1]=first[o];first[o]=e1++;}

int main()
{
	memset(first,-1,sizeof(first));
	scanf("%d%d",&n,&m);
	scanf("%s",ch);
	for(int i=0;i<n;i++) is[i+1]=ch[i]=='A';
	memset(a,0,sizeof(a));
	for(int i=1;i<=m;i++)
	{
		int x,y;scanf("%d%d",&x,&y);
		insert(x,y);insert(y,x);
		a[x][is[y]]++;a[y][is[x]]++;
	}
	memset(v,0,sizeof(v));
	for(int i=1;i<=n;i++) if((a[i][0]==0)||(a[i][1]==0)) q.push(i),v[i]=1,k++;
	while(!q.empty())
	{
		int yu=q.front();
		for(int i=first[yu];i!=-1;i=next[i])
		    if(!v[to[i]])
		    {
		    	a[to[i]][is[yu]]--;
		    	if(a[to[i]][is[yu]]==0) q.push(to[i]),v[to[i]]=1,k++;
			}
		q.pop();
	}
	if(k<n) cout<<"Yes"<<endl;else cout<<"No"<<endl;
	return 0;
} 
### Prompt

Construct a cpp code solution to the problem outlined:

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<iostream>
#include<cstring>
#include<cstdio>
#include<queue>
#define next sfhds

using namespace std;

int a[200001][2],n,m,to[500001],next[500001],first[500001],is[200001],e1=0,k=0,v[200001];
char ch[200001];
queue<int> q;

void insert(int o,int p) {to[e1]=p;next[e1]=first[o];first[o]=e1++;}

int main()
{
	memset(first,-1,sizeof(first));
	scanf("%d%d",&n,&m);
	scanf("%s",ch);
	for(int i=0;i<n;i++) is[i+1]=ch[i]=='A';
	memset(a,0,sizeof(a));
	for(int i=1;i<=m;i++)
	{
		int x,y;scanf("%d%d",&x,&y);
		insert(x,y);insert(y,x);
		a[x][is[y]]++;a[y][is[x]]++;
	}
	memset(v,0,sizeof(v));
	for(int i=1;i<=n;i++) if((a[i][0]==0)||(a[i][1]==0)) q.push(i),v[i]=1,k++;
	while(!q.empty())
	{
		int yu=q.front();
		for(int i=first[yu];i!=-1;i=next[i])
		    if(!v[to[i]])
		    {
		    	a[to[i]][is[yu]]--;
		    	if(a[to[i]][is[yu]]==0) q.push(to[i]),v[to[i]]=1,k++;
			}
		q.pop();
	}
	if(k<n) cout<<"Yes"<<endl;else cout<<"No"<<endl;
	return 0;
} 
```