    #include<bits/stdc++.h>
    #define LL long long
    #define clr(x,i) memset(x,i,sizeof(x))
    using namespace std;
    const int N=200005;
    int n,m,ind[N][2],vis[N];
    vector<int> G[N];
    char s[N];
    queue<int> q;
    int main()
    {
    	ios::sync_with_stdio(0);
    	cin>>n>>m>>s+1;
    	//cout<<s+1<<endl;
    	for(int i=1;i<=m;i++){
    		int u,v; cin>>u>>v;
    		G[u].push_back(v); G[v].push_back(u);
    		ind[u][s[v]-'A']++; ind[v][s[u]-'A']++;
    	}
    	int tot=0;
    	for(int i=1;i<=n;i++){
    		if(!ind[i][0]||!ind[i][1]) q.push(i),vis[i]=1;
    	}
    	while(!q.empty()){
    		int u=q.front(); q.pop(); tot++;
    		for(int i=0;i<G[u].size();i++){
    			int v=G[u][i];
    			if(!(--ind[v][s[u]-'A'])&&!vis[v]) q.push(v),vis[v]=1;
    		}
    	}
    	if(tot<n) puts("Yes");
    	else puts("No");
    	return 0;
    }
### Prompt

In CPP, your task is to solve 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>
    #define LL long long
    #define clr(x,i) memset(x,i,sizeof(x))
    using namespace std;
    const int N=200005;
    int n,m,ind[N][2],vis[N];
    vector<int> G[N];
    char s[N];
    queue<int> q;
    int main()
    {
    	ios::sync_with_stdio(0);
    	cin>>n>>m>>s+1;
    	//cout<<s+1<<endl;
    	for(int i=1;i<=m;i++){
    		int u,v; cin>>u>>v;
    		G[u].push_back(v); G[v].push_back(u);
    		ind[u][s[v]-'A']++; ind[v][s[u]-'A']++;
    	}
    	int tot=0;
    	for(int i=1;i<=n;i++){
    		if(!ind[i][0]||!ind[i][1]) q.push(i),vis[i]=1;
    	}
    	while(!q.empty()){
    		int u=q.front(); q.pop(); tot++;
    		for(int i=0;i<G[u].size();i++){
    			int v=G[u][i];
    			if(!(--ind[v][s[u]-'A'])&&!vis[v]) q.push(v),vis[v]=1;
    		}
    	}
    	if(tot<n) puts("Yes");
    	else puts("No");
    	return 0;
    }
```