#include<bits/stdc++.h>
typedef long long ll;
ll gi(){
	ll x=0,f=1;
	char ch=getchar();
	while(!isdigit(ch))f^=ch=='-',ch=getchar();
	while(isdigit(ch))x=x*10+ch-'0',ch=getchar();
	return f?x:-x;
}
std::vector<int>G[200010];
int d[200010][2],del[200010],fa[200010];
char s[200010];
std::queue<int>que;
int hd(int x){return fa[x]==x?x:fa[x]=hd(fa[x]);}
void Union(int x,int y){
	x=hd(x),y=hd(y);
	if(x==y)puts("Yes"),exit(0);
	fa[x]=y;
}
int main(){
#ifdef XZZSB
	freopen("in.in","r",stdin);
	freopen("out.out","w",stdout);
#endif
	int n=gi(),m=gi(),a,b;
	scanf("%s",s+1);
	while(m--)a=gi(),b=gi(),G[a].push_back(b),G[b].push_back(a),++d[a][s[b]-'A'],++d[b][s[a]-'A'];
	for(int i=1;i<=n;++i)if(!d[i][0]||!d[i][1])del[i]=1,que.push(i);
	while(!que.empty()){
		int x=que.front();que.pop();
		for(int i:G[x])if(!--d[i][s[x]-'A']&&!del[i])del[i]=1,que.push(i);
	}
	for(int i=1;i<=n;++i)if(!del[i])fa[i]=i;
	for(int i=1;i<=n;++i)if(!del[i])for(int j:G[i])if(!del[j]&&j>=i)Union(i,j);
	puts("No");
	return 0;
}


### Prompt

Your task is to create a cpp solution 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>
typedef long long ll;
ll gi(){
	ll x=0,f=1;
	char ch=getchar();
	while(!isdigit(ch))f^=ch=='-',ch=getchar();
	while(isdigit(ch))x=x*10+ch-'0',ch=getchar();
	return f?x:-x;
}
std::vector<int>G[200010];
int d[200010][2],del[200010],fa[200010];
char s[200010];
std::queue<int>que;
int hd(int x){return fa[x]==x?x:fa[x]=hd(fa[x]);}
void Union(int x,int y){
	x=hd(x),y=hd(y);
	if(x==y)puts("Yes"),exit(0);
	fa[x]=y;
}
int main(){
#ifdef XZZSB
	freopen("in.in","r",stdin);
	freopen("out.out","w",stdout);
#endif
	int n=gi(),m=gi(),a,b;
	scanf("%s",s+1);
	while(m--)a=gi(),b=gi(),G[a].push_back(b),G[b].push_back(a),++d[a][s[b]-'A'],++d[b][s[a]-'A'];
	for(int i=1;i<=n;++i)if(!d[i][0]||!d[i][1])del[i]=1,que.push(i);
	while(!que.empty()){
		int x=que.front();que.pop();
		for(int i:G[x])if(!--d[i][s[x]-'A']&&!del[i])del[i]=1,que.push(i);
	}
	for(int i=1;i<=n;++i)if(!del[i])fa[i]=i;
	for(int i=1;i<=n;++i)if(!del[i])for(int j:G[i])if(!del[j]&&j>=i)Union(i,j);
	puts("No");
	return 0;
}


```