#include <cstdio>
using namespace std;
const int N=200005;
int n,m,x,y,cnt,head[N],vis[N],deg[N][2],q[N],qtd,qtl;
char col[N];
struct edge{
  int to,nex;
}e[N*2];
void add(int x,int y){
  e[++cnt].to=y;e[cnt].nex=head[x];head[x]=cnt;
  e[++cnt].to=x;e[cnt].nex=head[y];head[y]=cnt;
}
int main(){
  scanf("%d%d",&n,&m);
  scanf("%s",col+1);
  for (int i=1;i<=m;i++) {
  scanf("%d%d",&x,&y),add(x,y);
  deg[x][col[y]=='B']++;deg[y][col[x]=='B']++;
  }
  for (int i=1;i<=n;i++)
    if (deg[i][0]==0||deg[i][1]==0) vis[q[qtl++]=i]=1;
  while (qtd!=qtl){
  	int x=q[qtd++];
  	for (int i=head[x];i;i=e[i].nex)
  	  if (!vis[e[i].to]){
  	  	 if (!--deg[e[i].to][col[x]=='B']){
  	  	 	  vis[e[i].to]=1;
  	  	 	  q[qtl++]=e[i].to;
			 }
		}
  }
  if (qtl==n) puts("No");
  else puts("Yes");
  return 0;
}
### 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 <cstdio>
using namespace std;
const int N=200005;
int n,m,x,y,cnt,head[N],vis[N],deg[N][2],q[N],qtd,qtl;
char col[N];
struct edge{
  int to,nex;
}e[N*2];
void add(int x,int y){
  e[++cnt].to=y;e[cnt].nex=head[x];head[x]=cnt;
  e[++cnt].to=x;e[cnt].nex=head[y];head[y]=cnt;
}
int main(){
  scanf("%d%d",&n,&m);
  scanf("%s",col+1);
  for (int i=1;i<=m;i++) {
  scanf("%d%d",&x,&y),add(x,y);
  deg[x][col[y]=='B']++;deg[y][col[x]=='B']++;
  }
  for (int i=1;i<=n;i++)
    if (deg[i][0]==0||deg[i][1]==0) vis[q[qtl++]=i]=1;
  while (qtd!=qtl){
  	int x=q[qtd++];
  	for (int i=head[x];i;i=e[i].nex)
  	  if (!vis[e[i].to]){
  	  	 if (!--deg[e[i].to][col[x]=='B']){
  	  	 	  vis[e[i].to]=1;
  	  	 	  q[qtl++]=e[i].to;
			 }
		}
  }
  if (qtl==n) puts("No");
  else puts("Yes");
  return 0;
}
```