#include<cstdio>
inline void read(int&r)
{
	register char c;r=0;
	do c=getchar();while(c<'0'||c>'9');
	do r=r*10+c-'0',c=getchar();while(c>='0'&&c<='9');
}

char s[200001];
int first[200001];
int a[200001],b[200001],q[400001];
struct ac{ int to,nx; }e[400001];
void readedge()
{
	static int L=1,u,v;read(u),read(v);
	e[L].to=v,e[L].nx=first[u],first[u]=L++;
	e[L].to=u,e[L].nx=first[v],first[v]=L++;
	(s[u]=='A'?a:b)[v]++;
	(s[v]=='A'?a:b)[u]++;
}

int main()
{
	int n,u,t=0,i,j;
	scanf("%d%d%s",&n,&t,s+1);
	while(t--)readedge();
	for(u=1,t=0;u<=n;u++)if(!a[u]||!b[u])q[t++]=u;
	for(i=0;i<t;i++)
		for(j=first[u=q[i]];j;j=e[j].nx)
			if(!--(s[u]=='A'?a:b)[e[j].to]&&(s[u]=='A'?b:a)[e[j].to])
				q[t++]=e[j].to;
	puts(t==n?"No":"Yes");
}
### Prompt

Construct a Cpp code solution to the problem outlined:

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>
inline void read(int&r)
{
	register char c;r=0;
	do c=getchar();while(c<'0'||c>'9');
	do r=r*10+c-'0',c=getchar();while(c>='0'&&c<='9');
}

char s[200001];
int first[200001];
int a[200001],b[200001],q[400001];
struct ac{ int to,nx; }e[400001];
void readedge()
{
	static int L=1,u,v;read(u),read(v);
	e[L].to=v,e[L].nx=first[u],first[u]=L++;
	e[L].to=u,e[L].nx=first[v],first[v]=L++;
	(s[u]=='A'?a:b)[v]++;
	(s[v]=='A'?a:b)[u]++;
}

int main()
{
	int n,u,t=0,i,j;
	scanf("%d%d%s",&n,&t,s+1);
	while(t--)readedge();
	for(u=1,t=0;u<=n;u++)if(!a[u]||!b[u])q[t++]=u;
	for(i=0;i<t;i++)
		for(j=first[u=q[i]];j;j=e[j].nx)
			if(!--(s[u]=='A'?a:b)[e[j].to]&&(s[u]=='A'?b:a)[e[j].to])
				q[t++]=e[j].to;
	puts(t==n?"No":"Yes");
}
```