#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
const int maxn=200010;
char col[maxn];
vector<int> g[maxn];
int acnt[maxn],bcnt[maxn];
bool vis[maxn];
int main(){
	int n,m,u,v;
	scanf("%d%d%s",&n,&m,col+1);
	for(int i=0;i<m;i++){
		scanf("%d%d",&u,&v);
		g[u].push_back(v);
		g[v].push_back(u);
	}
	queue<int> q;
	for(int u=1;u<=n;u++){
		for(int i=0;i<(int)g[u].size();i++){
			int v=g[u][i];
			if(col[v]=='A') acnt[u]++;
			else bcnt[u]++;
		}
		if(acnt[u]==0 || bcnt[u]==0){
			q.push(u);
			vis[u]=true;
		}
	}
	while(!q.empty()){
		int u=q.front();
		q.pop();
		for(int i=0;i<(int)g[u].size();i++){
			int v=g[u][i];
			if(!vis[v]){
				if(col[u]=='A') acnt[v]--;
				else bcnt[v]--;
				if(acnt[v]==0 || bcnt[v]==0){
					q.push(v);
					vis[v]=true;
				}
			}
		}
	}
	bool found=false;
	for(int v=1;v<=n;v++)
		if(!vis[v]) found=true;
	puts(found?"Yes":"No");
	return 0;
}
### Prompt

Generate 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<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
const int maxn=200010;
char col[maxn];
vector<int> g[maxn];
int acnt[maxn],bcnt[maxn];
bool vis[maxn];
int main(){
	int n,m,u,v;
	scanf("%d%d%s",&n,&m,col+1);
	for(int i=0;i<m;i++){
		scanf("%d%d",&u,&v);
		g[u].push_back(v);
		g[v].push_back(u);
	}
	queue<int> q;
	for(int u=1;u<=n;u++){
		for(int i=0;i<(int)g[u].size();i++){
			int v=g[u][i];
			if(col[v]=='A') acnt[u]++;
			else bcnt[u]++;
		}
		if(acnt[u]==0 || bcnt[u]==0){
			q.push(u);
			vis[u]=true;
		}
	}
	while(!q.empty()){
		int u=q.front();
		q.pop();
		for(int i=0;i<(int)g[u].size();i++){
			int v=g[u][i];
			if(!vis[v]){
				if(col[u]=='A') acnt[v]--;
				else bcnt[v]--;
				if(acnt[v]==0 || bcnt[v]==0){
					q.push(v);
					vis[v]=true;
				}
			}
		}
	}
	bool found=false;
	for(int v=1;v<=n;v++)
		if(!vis[v]) found=true;
	puts(found?"Yes":"No");
	return 0;
}
```