#include<bits/stdc++.h>
using namespace std;
const int maxn=200100;
char str[maxn];
int beg[maxn],tto[maxn<<1],nex[maxn<<1],e;
void putin(int s,int t){
	tto[++e]=t;
	nex[e]=beg[s];
	beg[s]=e;
}
int q[maxn],l,r;
bool inq[maxn];
int cntA[maxn],cntB[maxn];
int main(){
	int n,m;
	scanf("%d%d",&n,&m);
	scanf("%s",str+1);
	int s,t;
	for(int i=1;i<=m;i++){
		scanf("%d%d",&s,&t);
		putin(s,t);
		putin(t,s);
	}
	l=1,r=0;
	for(int i=1;i<=n;i++){
		for(int j=beg[i];j;j=nex[j]){
			if(str[tto[j]]=='A')
				cntA[i]++;
			else
				cntB[i]++;
		}
		if(cntA[i]==0||cntB[i]==0){
			q[++r]=i;
			inq[i]=1;
		}
	}
	int u;
	while(l<=r){
		u=q[l],l++;
		for(int i=beg[u];i;i=nex[i]){
			if(str[u]=='A')
				cntA[tto[i]]--;
			else
				cntB[tto[i]]--;
			if((cntA[tto[i]]==0||cntB[tto[i]]==0)&&!inq[tto[i]]){
				inq[tto[i]]=1;
				q[++r]=tto[i];
			}
		}
	}
	if(r<n)
		printf("Yes\n");
	else
		printf("No\n");
	return 0;
}
### Prompt

Generate 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>
using namespace std;
const int maxn=200100;
char str[maxn];
int beg[maxn],tto[maxn<<1],nex[maxn<<1],e;
void putin(int s,int t){
	tto[++e]=t;
	nex[e]=beg[s];
	beg[s]=e;
}
int q[maxn],l,r;
bool inq[maxn];
int cntA[maxn],cntB[maxn];
int main(){
	int n,m;
	scanf("%d%d",&n,&m);
	scanf("%s",str+1);
	int s,t;
	for(int i=1;i<=m;i++){
		scanf("%d%d",&s,&t);
		putin(s,t);
		putin(t,s);
	}
	l=1,r=0;
	for(int i=1;i<=n;i++){
		for(int j=beg[i];j;j=nex[j]){
			if(str[tto[j]]=='A')
				cntA[i]++;
			else
				cntB[i]++;
		}
		if(cntA[i]==0||cntB[i]==0){
			q[++r]=i;
			inq[i]=1;
		}
	}
	int u;
	while(l<=r){
		u=q[l],l++;
		for(int i=beg[u];i;i=nex[i]){
			if(str[u]=='A')
				cntA[tto[i]]--;
			else
				cntB[tto[i]]--;
			if((cntA[tto[i]]==0||cntB[tto[i]]==0)&&!inq[tto[i]]){
				inq[tto[i]]=1;
				q[++r]=tto[i];
			}
		}
	}
	if(r<n)
		printf("Yes\n");
	else
		printf("No\n");
	return 0;
}
```