#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<queue>
using namespace std;
const int N=200005,M=N*2;
int a[N];
int en=0,to[M],nxt[M],h[N];
void addedge(int u,int v)
	{
	to[++en]=v,nxt[en]=h[u],h[u]=en;
	}
int du[2][N];
int n,m;
int iff[N],cnt=0;
queue<int> q;

int main()
{
//freopen("input.txt","r",stdin);
cin >>n>>m;
char ch=getchar();
while(ch!='A'&&ch!='B')
	{
	ch=getchar();
	}
a[1]=ch=='A';
for(int i=2;i<=n;i++)
	{
	a[i]=getchar()=='A';
	}
for(int i=1;i<=m;i++)
	{
	int u,v;
	scanf("%d%d",&u,&v);
		du[a[v]][u]++;
		du[a[u]][v]++;
	addedge(u,v);
	addedge(v,u);
	}
for(int i=1;i<=n;i++)
	{
	if(!du[0][i]||!du[1][i])
		q.push(i),
		iff[i]=1,
		cnt++;
	}
while(!q.empty())
	{
	int u=q.front();q.pop();
	for(int i=h[u];i;i=nxt[i])
		{
		int v=to[i];
		du[a[u]][v]--;
		if(iff[v])continue;
		if(!du[a[u]][v]&&!iff[v])
			{
			q.push(v);
			iff[v]=1;
			cnt++;
			}
		}
	}
if(cnt<n)
	{
	cout <<"Yes";
	}
else cout <<"No";
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<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<queue>
using namespace std;
const int N=200005,M=N*2;
int a[N];
int en=0,to[M],nxt[M],h[N];
void addedge(int u,int v)
	{
	to[++en]=v,nxt[en]=h[u],h[u]=en;
	}
int du[2][N];
int n,m;
int iff[N],cnt=0;
queue<int> q;

int main()
{
//freopen("input.txt","r",stdin);
cin >>n>>m;
char ch=getchar();
while(ch!='A'&&ch!='B')
	{
	ch=getchar();
	}
a[1]=ch=='A';
for(int i=2;i<=n;i++)
	{
	a[i]=getchar()=='A';
	}
for(int i=1;i<=m;i++)
	{
	int u,v;
	scanf("%d%d",&u,&v);
		du[a[v]][u]++;
		du[a[u]][v]++;
	addedge(u,v);
	addedge(v,u);
	}
for(int i=1;i<=n;i++)
	{
	if(!du[0][i]||!du[1][i])
		q.push(i),
		iff[i]=1,
		cnt++;
	}
while(!q.empty())
	{
	int u=q.front();q.pop();
	for(int i=h[u];i;i=nxt[i])
		{
		int v=to[i];
		du[a[u]][v]--;
		if(iff[v])continue;
		if(!du[a[u]][v]&&!iff[v])
			{
			q.push(v);
			iff[v]=1;
			cnt++;
			}
		}
	}
if(cnt<n)
	{
	cout <<"Yes";
	}
else cout <<"No";
return 0;
}

```