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

string S;
vector<vector<int>> v(400010);
int N,M,visited[400010] = {0};

void dfs(int n){
	visited[n] = -1;
	for(int i=0;i<v[n].size();i++){
		if(visited[v[n][i]]==-1){
			cout << "Yes" << endl;
			exit(0);
		}
		else if(visited[v[n][i]]==0) dfs(v[n][i]);
	}
	visited[n] = 1;
}

int main(){
	cin  >> N >> M >> S;
	int a,b;
	for(int i=0;i<M;i++){
		cin >> a >> b;
		if(S[a-1]==S[b-1]){
			v[a].push_back(b+N);
			v[b].push_back(a+N);
		}else{
			v[a+N].push_back(b);
			v[b+N].push_back(a);
		}
	}
	for(int i=1;i<=N;i++){
		if(visited[i]==0) dfs(i);
	}
	cout << "No" << endl;
}
### 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 <iostream>
#include <vector>
#include <cstdlib>
using namespace std;

string S;
vector<vector<int>> v(400010);
int N,M,visited[400010] = {0};

void dfs(int n){
	visited[n] = -1;
	for(int i=0;i<v[n].size();i++){
		if(visited[v[n][i]]==-1){
			cout << "Yes" << endl;
			exit(0);
		}
		else if(visited[v[n][i]]==0) dfs(v[n][i]);
	}
	visited[n] = 1;
}

int main(){
	cin  >> N >> M >> S;
	int a,b;
	for(int i=0;i<M;i++){
		cin >> a >> b;
		if(S[a-1]==S[b-1]){
			v[a].push_back(b+N);
			v[b].push_back(a+N);
		}else{
			v[a+N].push_back(b);
			v[b+N].push_back(a);
		}
	}
	for(int i=1;i<=N;i++){
		if(visited[i]==0) dfs(i);
	}
	cout << "No" << endl;
}
```