#include<bits/stdc++.h>
using namespace std;

int cntA[200001],cntB[200001];
vector<int> e[200001];
bool vis[200001];
int main(){
	ios_base::sync_with_stdio(0);
	cin.tie(0);
	int n,m;
	string s;
	cin>>n>>m>>s;
	s.insert(0," ");
	for(int i=1,u,v;i<=m;i++){
		cin>>u>>v;
		e[u].emplace_back(v);
		e[v].emplace_back(u);
		if(s[v]=='A')
			cntA[u]++;
		else
			cntB[u]++;
		if(s[u]=='A')
			cntA[v]++;
		else
			cntB[v]++;
	}
	queue<int> q;
	int cnt = 0;
	for(int i=1;i<=n;i++)
		if(cntA[i]==0||cntB[i]==0)
			q.emplace(i),vis[i]=true;
	while(!q.empty()){
		int x = q.front();q.pop();
		cnt++;
		for(int it:e[x]){
			if(s[x]=='A')
				cntA[it]--;
			else
				cntB[it]--;
			if(!vis[it])
				if(cntA[it]==0||cntB[it]==0)
					q.emplace(it),vis[it]=true;
		}
	}
	if(cnt==n)
		cout<<"No"<<endl;
	else
		cout<<"Yes"<<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;

int cntA[200001],cntB[200001];
vector<int> e[200001];
bool vis[200001];
int main(){
	ios_base::sync_with_stdio(0);
	cin.tie(0);
	int n,m;
	string s;
	cin>>n>>m>>s;
	s.insert(0," ");
	for(int i=1,u,v;i<=m;i++){
		cin>>u>>v;
		e[u].emplace_back(v);
		e[v].emplace_back(u);
		if(s[v]=='A')
			cntA[u]++;
		else
			cntB[u]++;
		if(s[u]=='A')
			cntA[v]++;
		else
			cntB[v]++;
	}
	queue<int> q;
	int cnt = 0;
	for(int i=1;i<=n;i++)
		if(cntA[i]==0||cntB[i]==0)
			q.emplace(i),vis[i]=true;
	while(!q.empty()){
		int x = q.front();q.pop();
		cnt++;
		for(int it:e[x]){
			if(s[x]=='A')
				cntA[it]--;
			else
				cntB[it]--;
			if(!vis[it])
				if(cntA[it]==0||cntB[it]==0)
					q.emplace(it),vis[it]=true;
		}
	}
	if(cnt==n)
		cout<<"No"<<endl;
	else
		cout<<"Yes"<<endl;
}

```