#include<bits/stdc++.h>
#define LL long long

using namespace std;
const int N=2e5+10;
int rd()
{
    int x=0,w=1;char ch=0;
    while(ch<'0'||ch>'9'){if(ch=='-') w=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+(ch^48);ch=getchar();}
    return x*w;
}
int to[N<<1],nt[N<<1],hd[N],dg[N][2],tot=1;
void adde(int x,int y)
{
    ++tot,to[tot]=y,nt[tot]=hd[x],hd[x]=tot;
    ++tot,to[tot]=x,nt[tot]=hd[y],hd[y]=tot;
}
char cc[N];
int n,m,tt;
bool v[N];

int main()
{
    n=rd(),m=rd();
    scanf("%s",cc+1);
    for(int i=1;i<=m;++i)
    {
	int x=rd(),y=rd();
	adde(x,y);
	++dg[x][cc[y]-'A'],++dg[y][cc[x]-'A'];
    }
    queue<int> q;
    for(int i=1;i<=n;++i)
	if(!dg[i][0]||!dg[i][1]) v[i]=1,q.push(i);
    while(!q.empty())
    {
	int x=q.front();
	q.pop();
	++tt;
	for(int i=hd[x];i;i=nt[i])
	{
	    int y=to[i];
	    --dg[y][cc[x]-'A'];
	    if(!v[y]&&!dg[y][cc[x]-'A']) v[y]=1,q.push(y);
	}
    }
    puts(tt==n?"No":"Yes");
    return 0;
}



### Prompt

Please formulate 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>
#define LL long long

using namespace std;
const int N=2e5+10;
int rd()
{
    int x=0,w=1;char ch=0;
    while(ch<'0'||ch>'9'){if(ch=='-') w=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+(ch^48);ch=getchar();}
    return x*w;
}
int to[N<<1],nt[N<<1],hd[N],dg[N][2],tot=1;
void adde(int x,int y)
{
    ++tot,to[tot]=y,nt[tot]=hd[x],hd[x]=tot;
    ++tot,to[tot]=x,nt[tot]=hd[y],hd[y]=tot;
}
char cc[N];
int n,m,tt;
bool v[N];

int main()
{
    n=rd(),m=rd();
    scanf("%s",cc+1);
    for(int i=1;i<=m;++i)
    {
	int x=rd(),y=rd();
	adde(x,y);
	++dg[x][cc[y]-'A'],++dg[y][cc[x]-'A'];
    }
    queue<int> q;
    for(int i=1;i<=n;++i)
	if(!dg[i][0]||!dg[i][1]) v[i]=1,q.push(i);
    while(!q.empty())
    {
	int x=q.front();
	q.pop();
	++tt;
	for(int i=hd[x];i;i=nt[i])
	{
	    int y=to[i];
	    --dg[y][cc[x]-'A'];
	    if(!v[y]&&!dg[y][cc[x]-'A']) v[y]=1,q.push(y);
	}
    }
    puts(tt==n?"No":"Yes");
    return 0;
}



```