#include<bits/stdc++.h>
#define ll long long
using namespace std;
int tot,n,m,op[500010],e[500010],hd[500010],v[500010],du[500010][2],cnt,nt[500010];
char s[500010];
queue<int>q;
void build(int x,int y){
	tot++;
	e[tot]=y;
	nt[tot]=hd[x];hd[x]=tot;
}
int main(){
	int i,j,x,y;
	scanf("%d%d",&n,&m);
	scanf("%s",s+1);
	for(i=1;i<=n;i++)op[i]=(s[i]=='A'?0:1);
	for(i=1;i<=m;i++){
		scanf("%d%d",&x,&y);
		build(x,y);build(y,x);
		du[x][op[y]]++;du[y][op[x]]++;
	}
	for(i=1;i<=n;i++)if(!du[i][0]||!du[i][1]){
		cnt++;v[i]=1;q.push(i);
	}
	while(!q.empty()){
		x=q.front();q.pop();
		for(i=hd[x];i;i=nt[i])if(!v[e[i]]){
			du[e[i]][op[x]]--;
			if(!du[e[i]][op[x]]){
				cnt++;
				v[e[i]]=1;q.push(e[i]);
			}
		}
	}
	if(cnt==n)puts("No");
	 else puts("Yes");
}
### Prompt

In Cpp, your task is to solve 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 ll long long
using namespace std;
int tot,n,m,op[500010],e[500010],hd[500010],v[500010],du[500010][2],cnt,nt[500010];
char s[500010];
queue<int>q;
void build(int x,int y){
	tot++;
	e[tot]=y;
	nt[tot]=hd[x];hd[x]=tot;
}
int main(){
	int i,j,x,y;
	scanf("%d%d",&n,&m);
	scanf("%s",s+1);
	for(i=1;i<=n;i++)op[i]=(s[i]=='A'?0:1);
	for(i=1;i<=m;i++){
		scanf("%d%d",&x,&y);
		build(x,y);build(y,x);
		du[x][op[y]]++;du[y][op[x]]++;
	}
	for(i=1;i<=n;i++)if(!du[i][0]||!du[i][1]){
		cnt++;v[i]=1;q.push(i);
	}
	while(!q.empty()){
		x=q.front();q.pop();
		for(i=hd[x];i;i=nt[i])if(!v[e[i]]){
			du[e[i]][op[x]]--;
			if(!du[e[i]][op[x]]){
				cnt++;
				v[e[i]]=1;q.push(e[i]);
			}
		}
	}
	if(cnt==n)puts("No");
	 else puts("Yes");
}
```