#include<iostream>
#include<vector>
using namespace std;
char ca[200005];
vector<int> path[200005];
int nnext[200005][2];
bool dead[200005];

void rem(int a){
	int i,j,k;
	int t;
	for(i=0;i<path[a].size();i++){
		t=path[a][i];
		if(dead[t])continue;
		if(ca[a]=='A')nnext[t][0]--;
		else nnext[t][1]--;
		if(nnext[t][0]==0 || nnext[t][1]==0){
			dead[t]=true;
			rem(t);
		}
	}
}
		
bool check(int a){
	if(dead[a])return false;
	for(int i=0;i<path[a].size();i++){
		if(!dead[path[a][i]])return true;
	}
	return false;
}

int main(){
	int n,m;
	int a,b,c;
	int i,j,k;
	bool s=false;
	cin>>n>>m;
	cin>>ca;
	for(i=0;i<m;i++){
		cin>>a>>b;
		a--,b--;
		path[a].push_back(b);
		path[b].push_back(a);
		if(ca[a]=='A')nnext[b][0]++;
		else nnext[b][1]++;
		if(ca[b]=='A')nnext[a][0]++;
		else nnext[a][1]++;
	}
	for(i=0;i<n;i++){
		if(dead[i])continue;
		if(nnext[i][0]==0 || nnext[i][1]==0){
			dead[i]=true;
			rem(i);
		}
	}
	for(i=0;i<n;i++){
		if(check(i))s=true;
	}
	if(s)cout<<"Yes"<<endl;
	else cout<<"No"<<endl;
	return 0;
}
	

### Prompt

Construct a cpp code solution to the problem outlined:

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>
using namespace std;
char ca[200005];
vector<int> path[200005];
int nnext[200005][2];
bool dead[200005];

void rem(int a){
	int i,j,k;
	int t;
	for(i=0;i<path[a].size();i++){
		t=path[a][i];
		if(dead[t])continue;
		if(ca[a]=='A')nnext[t][0]--;
		else nnext[t][1]--;
		if(nnext[t][0]==0 || nnext[t][1]==0){
			dead[t]=true;
			rem(t);
		}
	}
}
		
bool check(int a){
	if(dead[a])return false;
	for(int i=0;i<path[a].size();i++){
		if(!dead[path[a][i]])return true;
	}
	return false;
}

int main(){
	int n,m;
	int a,b,c;
	int i,j,k;
	bool s=false;
	cin>>n>>m;
	cin>>ca;
	for(i=0;i<m;i++){
		cin>>a>>b;
		a--,b--;
		path[a].push_back(b);
		path[b].push_back(a);
		if(ca[a]=='A')nnext[b][0]++;
		else nnext[b][1]++;
		if(ca[b]=='A')nnext[a][0]++;
		else nnext[a][1]++;
	}
	for(i=0;i<n;i++){
		if(dead[i])continue;
		if(nnext[i][0]==0 || nnext[i][1]==0){
			dead[i]=true;
			rem(i);
		}
	}
	for(i=0;i<n;i++){
		if(check(i))s=true;
	}
	if(s)cout<<"Yes"<<endl;
	else cout<<"No"<<endl;
	return 0;
}
	

```