#include<bits/stdc++.h>

using namespace std;
#define NMAX 200045
vector<int> g[NMAX];
int a[NMAX][2];
bool s[NMAX];
signed main(){
	int n, m; cin>>n>>m;
	for(int i= 0; i < n; i++){
		char t; cin>>t;
		s[i] = (t == 'B');
	}
	
	for(int i = 0; i < m; i++){
		int a, b; cin>>a>>b; a--,b --;
		g[a].push_back(b);
		g[b].push_back(a);
	}
	
	queue<int> q;
	vector<bool> alive(n);
	for(int i = 0; i < n; i++){
		//cout<<i<<endl;
		for(int adj: g[i]){
			a[i][s[adj]]++;
		}
		if(a[i][0] == 0 || a[i][1] == 0){
			q.push(i);
			alive[i] = true;
		}
	}
	while(!q.empty()){
		int cur = q.front(); q.pop();
		
		for(int adj: g[cur]){
			if(!alive[adj]){
				a[adj][s[cur]]--;
				if(a[adj][0] <= 0 || a[adj][1] <= 0){
					alive[adj] = true;
					q.push(adj);
				}
			}
		}
	}
	int count = 0;
	for(int i = 0; i < n; i++){
		if(alive[i] == false){
			count++;
		}
	}
	if(count) cout<<"Yes"<<endl;
	else cout<<"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<bits/stdc++.h>

using namespace std;
#define NMAX 200045
vector<int> g[NMAX];
int a[NMAX][2];
bool s[NMAX];
signed main(){
	int n, m; cin>>n>>m;
	for(int i= 0; i < n; i++){
		char t; cin>>t;
		s[i] = (t == 'B');
	}
	
	for(int i = 0; i < m; i++){
		int a, b; cin>>a>>b; a--,b --;
		g[a].push_back(b);
		g[b].push_back(a);
	}
	
	queue<int> q;
	vector<bool> alive(n);
	for(int i = 0; i < n; i++){
		//cout<<i<<endl;
		for(int adj: g[i]){
			a[i][s[adj]]++;
		}
		if(a[i][0] == 0 || a[i][1] == 0){
			q.push(i);
			alive[i] = true;
		}
	}
	while(!q.empty()){
		int cur = q.front(); q.pop();
		
		for(int adj: g[cur]){
			if(!alive[adj]){
				a[adj][s[cur]]--;
				if(a[adj][0] <= 0 || a[adj][1] <= 0){
					alive[adj] = true;
					q.push(adj);
				}
			}
		}
	}
	int count = 0;
	for(int i = 0; i < n; i++){
		if(alive[i] == false){
			count++;
		}
	}
	if(count) cout<<"Yes"<<endl;
	else cout<<"No"<<endl;
}

```