#include <bits/stdc++.h>
using namespace std;
#define pb push_back
#define mp make_pair
#define fi first
#define se second
int n,m,rem[200007],u,v,col[200007],cnt[2][200007];
vector <int> edge[200007];
string s;
queue <int> q;
int main(){
	ios_base::sync_with_stdio(0);
	cin>>n>>m;
	cin>>s;
	for(int i=1;i<=n;i++) col[i]=s[i-1]-'A';
	for(int i=1;i<=m;i++){
		cin>>u>>v;
		edge[u].pb(v);
		edge[v].pb(u);
	}
	for(int i=1;i<=n;i++){
		for(int v:edge[i]) cnt[col[v]][i]++;
		//cout<<cnt[0][i]<<" "<<cnt[1][i]<<endl;
		if(cnt[0][i]==0 || cnt[1][i]==0) rem[i]=1, q.push(i);
	}
	while(!q.empty()){
		int u=q.front();
		//cout<<u<<endl;
		q.pop();
		for(int v:edge[u]){
			if(rem[v]==1) continue;
			cnt[col[u]][v]--;
			if(cnt[col[u]][v]==0) rem[v]=1, q.push(v);
		}
	}
	for(int i=1;i<=n;i++){
		if(rem[i]==0) return cout<<"Yes", 0;
	}
	cout<<"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 <bits/stdc++.h>
using namespace std;
#define pb push_back
#define mp make_pair
#define fi first
#define se second
int n,m,rem[200007],u,v,col[200007],cnt[2][200007];
vector <int> edge[200007];
string s;
queue <int> q;
int main(){
	ios_base::sync_with_stdio(0);
	cin>>n>>m;
	cin>>s;
	for(int i=1;i<=n;i++) col[i]=s[i-1]-'A';
	for(int i=1;i<=m;i++){
		cin>>u>>v;
		edge[u].pb(v);
		edge[v].pb(u);
	}
	for(int i=1;i<=n;i++){
		for(int v:edge[i]) cnt[col[v]][i]++;
		//cout<<cnt[0][i]<<" "<<cnt[1][i]<<endl;
		if(cnt[0][i]==0 || cnt[1][i]==0) rem[i]=1, q.push(i);
	}
	while(!q.empty()){
		int u=q.front();
		//cout<<u<<endl;
		q.pop();
		for(int v:edge[u]){
			if(rem[v]==1) continue;
			cnt[col[u]][v]--;
			if(cnt[col[u]][v]==0) rem[v]=1, q.push(v);
		}
	}
	for(int i=1;i<=n;i++){
		if(rem[i]==0) return cout<<"Yes", 0;
	}
	cout<<"No";
}

```