#include<iostream>
#include<vector>
using namespace std;
int N,M;
string s;
vector<int>G[2<<17];
vector<int>H[4<<17];
int color[4<<17];
bool dfs(int u,int p)
{
	color[u]=1;
	for(int v:H[u])
	{
		if(p==v)continue;
		else if(color[v]==1)return true;
		else if(color[v]==0&&dfs(v,u))return true;
	}
	color[u]=2;
	return false;
}
int main()
{
	cin>>N>>M>>s;
	for(int i=0;i<M;i++)
	{
		int a,b;cin>>a>>b;
		a--,b--;
		G[a].push_back(b);
		G[b].push_back(a);
	}
	for(int i=0;i<N;i++)
	{
		for(int v:G[i])
		{
			if(s[i]=='A')
			{
				if(s[v]=='B')H[i*2].push_back(v*2);
				else H[i*2+1].push_back(v*2);
			}
			else
			{
				if(s[v]=='B')H[i*2].push_back(v*2+1);
				else H[i*2+1].push_back(v*2+1);
			}
		}
	}
	for(int i=0;i<N*2;i++)
	{
		if(color[i]==0)
		{
			if(dfs(i,-1))
			{
				cout<<"Yes"<<endl;
				return 0;
			}
		}
	}
	cout<<"No"<<endl;
}
### Prompt

Your task is to create 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<iostream>
#include<vector>
using namespace std;
int N,M;
string s;
vector<int>G[2<<17];
vector<int>H[4<<17];
int color[4<<17];
bool dfs(int u,int p)
{
	color[u]=1;
	for(int v:H[u])
	{
		if(p==v)continue;
		else if(color[v]==1)return true;
		else if(color[v]==0&&dfs(v,u))return true;
	}
	color[u]=2;
	return false;
}
int main()
{
	cin>>N>>M>>s;
	for(int i=0;i<M;i++)
	{
		int a,b;cin>>a>>b;
		a--,b--;
		G[a].push_back(b);
		G[b].push_back(a);
	}
	for(int i=0;i<N;i++)
	{
		for(int v:G[i])
		{
			if(s[i]=='A')
			{
				if(s[v]=='B')H[i*2].push_back(v*2);
				else H[i*2+1].push_back(v*2);
			}
			else
			{
				if(s[v]=='B')H[i*2].push_back(v*2+1);
				else H[i*2+1].push_back(v*2+1);
			}
		}
	}
	for(int i=0;i<N*2;i++)
	{
		if(color[i]==0)
		{
			if(dfs(i,-1))
			{
				cout<<"Yes"<<endl;
				return 0;
			}
		}
	}
	cout<<"No"<<endl;
}
```