#include<iostream>
#include<cstdlib>
#include<algorithm>
#include<cstdio>
#include<cstring> 
#include<vector>
#include<queue>
using namespace std;
const int N=200100;
int n,m,val[N],vis[N];
vector<int> E[N];
char s[N];
queue<int> Q;
int check(int i)
{
	int v1=0,v2=0;			
	for(auto x:E[i])
	{
		if(vis[x]) continue;
		if(val[x]==0) v1=1;
		else v2=1;
	}
	return v1*v2;
}
int main()
{
	cin>>n;cin>>m;
	scanf("%s",s+1);
	for(int i=1;i<=n;i++) val[i]=s[i]-'A';
	for(int i=1;i<=m;i++)
	{
		int x,y;cin>>x>>y;
		E[x].push_back(y);
		if(x^y) E[y].push_back(x);	
	}
	for(int i=1;i<=n;i++) 
		if(!check(i)) Q.push(i),vis[i]=1;
	while(!Q.empty())
	{
		int x=Q.front();Q.pop();
		for(auto p:E[x]) 
			if(!vis[p]&&!check(p))
				Q.push(p),vis[p]=1; 
	}
	for(int i=1;i<=n;i++)
		if(!vis[i]) {puts("Yes");return 0;}
	puts("No");
}
### 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<cstdlib>
#include<algorithm>
#include<cstdio>
#include<cstring> 
#include<vector>
#include<queue>
using namespace std;
const int N=200100;
int n,m,val[N],vis[N];
vector<int> E[N];
char s[N];
queue<int> Q;
int check(int i)
{
	int v1=0,v2=0;			
	for(auto x:E[i])
	{
		if(vis[x]) continue;
		if(val[x]==0) v1=1;
		else v2=1;
	}
	return v1*v2;
}
int main()
{
	cin>>n;cin>>m;
	scanf("%s",s+1);
	for(int i=1;i<=n;i++) val[i]=s[i]-'A';
	for(int i=1;i<=m;i++)
	{
		int x,y;cin>>x>>y;
		E[x].push_back(y);
		if(x^y) E[y].push_back(x);	
	}
	for(int i=1;i<=n;i++) 
		if(!check(i)) Q.push(i),vis[i]=1;
	while(!Q.empty())
	{
		int x=Q.front();Q.pop();
		for(auto p:E[x]) 
			if(!vis[p]&&!check(p))
				Q.push(p),vis[p]=1; 
	}
	for(int i=1;i<=n;i++)
		if(!vis[i]) {puts("Yes");return 0;}
	puts("No");
}
```