#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#define MN 200005
using namespace std;
int n,m,num,que[MN<<4],head[MN<<4],rd[MN<<4];char s[MN];
struct edge{int to,next;}g[MN<<4];
void ins(int u,int v){g[++num].next=head[u];head[u]=num;g[num].to=v;}
int main(){
	scanf("%d%d%s",&n,&m,s+1);int x,y;
	for(int i=1;i<=m;i++){
		scanf("%d%d",&x,&y);
		if(x==y)ins(x+n,x+2*n),ins(x+3*n,x);
		else if(s[x]==s[y])ins(x+n,y),ins(x+n,y+2*n),ins(x+3*n,y),ins(x+3*n,y+2*n),ins(y+n,x),ins(y+n,x+2*n),ins(y+3*n,x),ins(y+3*n,x+2*n);
		else ins(x,y+n),ins(x,y+3*n),ins(x+2*n,y+n),ins(x+2*n,y+3*n),ins(y,x+n),ins(y,x+3*n),ins(y+2*n,x+n),ins(y+2*n,x+3*n);
	}
	for(int i=1;i<=4*n;i++)for(int j=head[i];j;j=g[j].next)rd[g[j].to]++;
	int h=1,t=0;for(int i=1;i<=4*n;i++)if(rd[i]==0)que[++t]=i;
	while(h<=t){
		
		int tmp=que[h++];
		for(int i=head[tmp];i;i=g[i].next)if(!(--rd[g[i].to]))que[++t]=g[i].to;
	}if(t==n*4)puts("No");else puts("Yes");
}

### 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<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#define MN 200005
using namespace std;
int n,m,num,que[MN<<4],head[MN<<4],rd[MN<<4];char s[MN];
struct edge{int to,next;}g[MN<<4];
void ins(int u,int v){g[++num].next=head[u];head[u]=num;g[num].to=v;}
int main(){
	scanf("%d%d%s",&n,&m,s+1);int x,y;
	for(int i=1;i<=m;i++){
		scanf("%d%d",&x,&y);
		if(x==y)ins(x+n,x+2*n),ins(x+3*n,x);
		else if(s[x]==s[y])ins(x+n,y),ins(x+n,y+2*n),ins(x+3*n,y),ins(x+3*n,y+2*n),ins(y+n,x),ins(y+n,x+2*n),ins(y+3*n,x),ins(y+3*n,x+2*n);
		else ins(x,y+n),ins(x,y+3*n),ins(x+2*n,y+n),ins(x+2*n,y+3*n),ins(y,x+n),ins(y,x+3*n),ins(y+2*n,x+n),ins(y+2*n,x+3*n);
	}
	for(int i=1;i<=4*n;i++)for(int j=head[i];j;j=g[j].next)rd[g[j].to]++;
	int h=1,t=0;for(int i=1;i<=4*n;i++)if(rd[i]==0)que[++t]=i;
	while(h<=t){
		
		int tmp=que[h++];
		for(int i=head[tmp];i;i=g[i].next)if(!(--rd[g[i].to]))que[++t]=g[i].to;
	}if(t==n*4)puts("No");else puts("Yes");
}

```