# include <bits/stdc++.h>
using namespace std;
int n,m,x,y,A[200005],B[200005],k,t,f[200005];
vector <int>v[200005];
string s;
queue<int>q;
int main (){
	cin>>n>>m>>s;
	for (int i=0; i<m; i++){
		cin>>x>>y; x--; y--;
		if (x==y)if(s[x]=='A')A[x]=1000000;else B[x]=1000000;
		else{
			if (s[x]=='A') A[y]++; else B[y]++;
			if (s[y]=='A') A[x]++; else B[x]++;
			v[x].push_back(y);
			v[y].push_back(x);
		}
	}
	for (int i=0; i<n; i++)if (A[i]*B[i]==0)q.push(i),f[i]=1;
	
	while(q.size()){
		k=q.front();
		q.pop();
		if (s[k]=='A') 
			for (int i=0; i<v[k].size(); i++){
				A[v[k][i]]--; 
				if (A[v[k][i]]<=0 && f[v[k][i]]==0)q.push(v[k][i]),f[v[k][i]]=1;
				}
		else	
			for (int i=0; i<v[k].size(); i++){
				B[v[k][i]]--; 
				if (B[v[k][i]]<=0 && f[v[k][i]]==0)q.push(v[k][i]),f[v[k][i]]=1;
				}	
	}
	for (int i=0; i<n; i++)if(A[i]>0 && B[i]>0){cout<<"Yes";return 0;}
	cout<<"No";
	return 0;
}
### Prompt

Please create a solution in cpp 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>
using namespace std;
int n,m,x,y,A[200005],B[200005],k,t,f[200005];
vector <int>v[200005];
string s;
queue<int>q;
int main (){
	cin>>n>>m>>s;
	for (int i=0; i<m; i++){
		cin>>x>>y; x--; y--;
		if (x==y)if(s[x]=='A')A[x]=1000000;else B[x]=1000000;
		else{
			if (s[x]=='A') A[y]++; else B[y]++;
			if (s[y]=='A') A[x]++; else B[x]++;
			v[x].push_back(y);
			v[y].push_back(x);
		}
	}
	for (int i=0; i<n; i++)if (A[i]*B[i]==0)q.push(i),f[i]=1;
	
	while(q.size()){
		k=q.front();
		q.pop();
		if (s[k]=='A') 
			for (int i=0; i<v[k].size(); i++){
				A[v[k][i]]--; 
				if (A[v[k][i]]<=0 && f[v[k][i]]==0)q.push(v[k][i]),f[v[k][i]]=1;
				}
		else	
			for (int i=0; i<v[k].size(); i++){
				B[v[k][i]]--; 
				if (B[v[k][i]]<=0 && f[v[k][i]]==0)q.push(v[k][i]),f[v[k][i]]=1;
				}	
	}
	for (int i=0; i<n; i++)if(A[i]>0 && B[i]>0){cout<<"Yes";return 0;}
	cout<<"No";
	return 0;
}
```