#include <iostream>
#include <vector>
#include <queue>
using namespace std;

string S;
vector<vector<int>> v(200010);
int N,M,visited[200010] = {0};

int main(){
	cin  >> N >> M >> S;
	int a,b;
	int A[200010] = {0},B[200010] = {0};
	for(int i=0;i<M;i++){
		cin >> a >> b;
		v[a].push_back(b);
		v[b].push_back(a);
	}
	queue<int> Q;
	int count = 0;
	for(int i=1;i<=N;i++){
		for(int j=0;j<v[i].size();j++){
			if(S[v[i][j]-1]=='A') A[i]++;
			if(S[v[i][j]-1]=='B') B[i]++;
		}
		if(A[i]==0 || B[i]==0){
			Q.push(i);
			visited[i] = 1;
			count++;
		}
	}
	while(!Q.empty()){
		int t = Q.front();
		Q.pop();
		for(int i=0;i<v[t].size();i++){
			if(S[t-1]=='A') A[v[t][i]]--;
			if(S[t-1]=='B') B[v[t][i]]--;
			if((A[v[t][i]]==0 || B[v[t][i]]==0) && visited[v[t][i]]==0){
				Q.push(v[t][i]);
				count++;
				visited[v[t][i]] = 1;
			}
		}
	}
	cout << (count<N? "Yes":"No") << endl;
}
### 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 <iostream>
#include <vector>
#include <queue>
using namespace std;

string S;
vector<vector<int>> v(200010);
int N,M,visited[200010] = {0};

int main(){
	cin  >> N >> M >> S;
	int a,b;
	int A[200010] = {0},B[200010] = {0};
	for(int i=0;i<M;i++){
		cin >> a >> b;
		v[a].push_back(b);
		v[b].push_back(a);
	}
	queue<int> Q;
	int count = 0;
	for(int i=1;i<=N;i++){
		for(int j=0;j<v[i].size();j++){
			if(S[v[i][j]-1]=='A') A[i]++;
			if(S[v[i][j]-1]=='B') B[i]++;
		}
		if(A[i]==0 || B[i]==0){
			Q.push(i);
			visited[i] = 1;
			count++;
		}
	}
	while(!Q.empty()){
		int t = Q.front();
		Q.pop();
		for(int i=0;i<v[t].size();i++){
			if(S[t-1]=='A') A[v[t][i]]--;
			if(S[t-1]=='B') B[v[t][i]]--;
			if((A[v[t][i]]==0 || B[v[t][i]]==0) && visited[v[t][i]]==0){
				Q.push(v[t][i]);
				count++;
				visited[v[t][i]] = 1;
			}
		}
	}
	cout << (count<N? "Yes":"No") << endl;
}
```