#include <iostream>
#include <stdio.h>
#include <string.h>
#include <vector>
#include <queue>
#include <algorithm>
using namespace std;
#define  LL long long
const int N = 2e5+33;
vector<int>G[N];
char k[N];
int d[N][2];
queue<int>q;
int vis[N];
int main()
{
	int n,m;
	scanf("%d%d",&n,&m);
	scanf("%s",k+1);
	for(int i=1;i<=m;i++){
		int u,v;
		scanf("%d%d",&u,&v);
		G[u].push_back(v);
		G[v].push_back(u);
		d[u][k[v]-'A']++;
		d[v][k[u]-'A']++;
	}
	for(int i=1;i<=n;i++) if(d[i][0]==0||d[i][1]==0) q.push(i),vis[i]=1;
	int cnt=(int)q.size();
	while(!q.empty()){
		int u=q.front();
		q.pop();
		for(int i=0;i<G[u].size();i++){
			int v=G[u][i];
			if(--d[v][k[u]-'A']==0&&vis[v]==0){
				q.push(v);
				cnt++;
				vis[v]=1;
			}
		}
	}
	if(cnt!=n){
		puts("Yes");
	}else{
		puts("No");
	}


	
	return 0;
}

### Prompt

Create a solution in CPP for 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 <stdio.h>
#include <string.h>
#include <vector>
#include <queue>
#include <algorithm>
using namespace std;
#define  LL long long
const int N = 2e5+33;
vector<int>G[N];
char k[N];
int d[N][2];
queue<int>q;
int vis[N];
int main()
{
	int n,m;
	scanf("%d%d",&n,&m);
	scanf("%s",k+1);
	for(int i=1;i<=m;i++){
		int u,v;
		scanf("%d%d",&u,&v);
		G[u].push_back(v);
		G[v].push_back(u);
		d[u][k[v]-'A']++;
		d[v][k[u]-'A']++;
	}
	for(int i=1;i<=n;i++) if(d[i][0]==0||d[i][1]==0) q.push(i),vis[i]=1;
	int cnt=(int)q.size();
	while(!q.empty()){
		int u=q.front();
		q.pop();
		for(int i=0;i<G[u].size();i++){
			int v=G[u][i];
			if(--d[v][k[u]-'A']==0&&vis[v]==0){
				q.push(v);
				cnt++;
				vis[v]=1;
			}
		}
	}
	if(cnt!=n){
		puts("Yes");
	}else{
		puts("No");
	}


	
	return 0;
}

```