#include <bits/stdc++.h>
#define F first
#define S second
using namespace std;

const int N=2e5+10;
vector<int>g[N];
int main(){
	int n, m;
	cin>>n>>m;
	string s;
	cin>>s;
	vector<int>a(n);
	vector<int>b(n);
	
	int u, v;
	for(int i=0;i<m;i++){
		cin>>u>>v;
		u--,v--;
		g[u].push_back(v);
		g[v].push_back(u);
		if(s[u]==s[v]){
			if(s[u]=='A')a[u]++,a[v]++;
			else b[u]++,b[v]++;
		}
		else{
			if(s[u]=='A')b[u]++,a[v]++;
			else a[u]++,b[v]++;
		}
	}
	queue<int>q;
	vector<int>d(n);
	for(int i=0;i<n;i++)
		if(!a[i]||!b[i]){
			q.push(i);
			d[i]=1;
		}
	while(q.size()){
		int u=q.front();q.pop();
		for(auto v:g[u])
			if(!d[v]){
				if(s[u]=='A')a[v]--;
				else b[v]--;
				if(!a[v]||!b[v]){
					d[v]=1;
					q.push(v);
				}
			}
	}
	for(int i=0;i<n;i++)
		if(!d[i]){
			cout<<"Yes\n";
			return 0;
		}
	cout<<"No\n";
	
}



### Prompt

Create a solution in CPP for 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 F first
#define S second
using namespace std;

const int N=2e5+10;
vector<int>g[N];
int main(){
	int n, m;
	cin>>n>>m;
	string s;
	cin>>s;
	vector<int>a(n);
	vector<int>b(n);
	
	int u, v;
	for(int i=0;i<m;i++){
		cin>>u>>v;
		u--,v--;
		g[u].push_back(v);
		g[v].push_back(u);
		if(s[u]==s[v]){
			if(s[u]=='A')a[u]++,a[v]++;
			else b[u]++,b[v]++;
		}
		else{
			if(s[u]=='A')b[u]++,a[v]++;
			else a[u]++,b[v]++;
		}
	}
	queue<int>q;
	vector<int>d(n);
	for(int i=0;i<n;i++)
		if(!a[i]||!b[i]){
			q.push(i);
			d[i]=1;
		}
	while(q.size()){
		int u=q.front();q.pop();
		for(auto v:g[u])
			if(!d[v]){
				if(s[u]=='A')a[v]--;
				else b[v]--;
				if(!a[v]||!b[v]){
					d[v]=1;
					q.push(v);
				}
			}
	}
	for(int i=0;i<n;i++)
		if(!d[i]){
			cout<<"Yes\n";
			return 0;
		}
	cout<<"No\n";
	
}



```