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

const int maxn=2e5+10;
int n,m,cnt,a[maxn<<1],b[maxn<<1],t[maxn<<1],xx;
char s[maxn];
vector<pair<int,int> > g[4][maxn];
int vis[maxn<<1];

inline void ae(int u,int v){
	++cnt;
	a[cnt]=u;b[cnt]=v;
	if(s[u]=='A')
		if(s[v]=='A')
			g[0][u].push_back(make_pair(v,cnt)),t[cnt]=0;
		else
			g[1][u].push_back(make_pair(v,cnt)),t[cnt]=1;
	else
		if(s[v]=='A')
			g[3][u].push_back(make_pair(v,cnt)),t[cnt]=3;
		else
			g[2][u].push_back(make_pair(v,cnt)),t[cnt]=2;
}
void dfs(int id){
	if(!vis[id])
		vis[id]=1;
	else if(vis[id]!=1)
		return;
	else{
		puts("Yes");
		exit(0);
	}
	for(int i=0,nxt=(t[id]+1)%4;i<g[nxt][b[id]].size();++i)
		dfs(g[nxt][b[id]][i].second);
	vis[id]=-1;
}

int main(){
	cin>>n>>m>>(s+1);
	for(int i=1,u,v;i<=m;++i){
		cin>>u>>v;
		ae(u,v);ae(v,u);
	}
	for(int i=1;i<=cnt;++i)
		if(!vis[i])
			dfs(i);
	puts("No");
	return 0;
}
### Prompt

Develop a solution in CPP to the problem described below:

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;

const int maxn=2e5+10;
int n,m,cnt,a[maxn<<1],b[maxn<<1],t[maxn<<1],xx;
char s[maxn];
vector<pair<int,int> > g[4][maxn];
int vis[maxn<<1];

inline void ae(int u,int v){
	++cnt;
	a[cnt]=u;b[cnt]=v;
	if(s[u]=='A')
		if(s[v]=='A')
			g[0][u].push_back(make_pair(v,cnt)),t[cnt]=0;
		else
			g[1][u].push_back(make_pair(v,cnt)),t[cnt]=1;
	else
		if(s[v]=='A')
			g[3][u].push_back(make_pair(v,cnt)),t[cnt]=3;
		else
			g[2][u].push_back(make_pair(v,cnt)),t[cnt]=2;
}
void dfs(int id){
	if(!vis[id])
		vis[id]=1;
	else if(vis[id]!=1)
		return;
	else{
		puts("Yes");
		exit(0);
	}
	for(int i=0,nxt=(t[id]+1)%4;i<g[nxt][b[id]].size();++i)
		dfs(g[nxt][b[id]][i].second);
	vis[id]=-1;
}

int main(){
	cin>>n>>m>>(s+1);
	for(int i=1,u,v;i<=m;++i){
		cin>>u>>v;
		ae(u,v);ae(v,u);
	}
	for(int i=1;i<=cnt;++i)
		if(!vis[i])
			dfs(i);
	puts("No");
	return 0;
}
```