#include<bits/stdc++.h>
using namespace std;
inline int read()
{
	int x;char c;
	while((c=getchar())<'0'||c>'9');
	for(x=c-'0';(c=getchar())>='0'&&c<='9';)x=x*10+c-'0';
	return x;
}
#define MN 200000
struct edge{int nx,t;}e[MN*2+5];
int h[MN+5],en,f[MN+5][2],q[MN+5],qn;
char s[MN+5];
inline void ins(int x,int y)
{
	e[++en]=(edge){h[x],y};h[x]=en;
	e[++en]=(edge){h[y],x};h[y]=en;
}
int main()
{
	int n,m,i,j;
	n=read();m=read();
	scanf("%s",s+1);
	while(m--)ins(read(),read());
	for(i=1;i<=n;++i)for(j=h[i];j;j=e[j].nx)++f[i][s[e[j].t]-'A'];
	for(i=1;i<=n;++i)if(!f[i][0]||!f[i][1])q[++qn]=i;
	for(i=1;i<=qn;++i)for(j=h[q[i]];j;j=e[j].nx)
		if(f[e[j].t][0]&&f[e[j].t][1]&&!--f[e[j].t][s[q[i]]-'A'])q[++qn]=e[j].t;
	puts(qn<n?"Yes":"No");
}
### 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<bits/stdc++.h>
using namespace std;
inline int read()
{
	int x;char c;
	while((c=getchar())<'0'||c>'9');
	for(x=c-'0';(c=getchar())>='0'&&c<='9';)x=x*10+c-'0';
	return x;
}
#define MN 200000
struct edge{int nx,t;}e[MN*2+5];
int h[MN+5],en,f[MN+5][2],q[MN+5],qn;
char s[MN+5];
inline void ins(int x,int y)
{
	e[++en]=(edge){h[x],y};h[x]=en;
	e[++en]=(edge){h[y],x};h[y]=en;
}
int main()
{
	int n,m,i,j;
	n=read();m=read();
	scanf("%s",s+1);
	while(m--)ins(read(),read());
	for(i=1;i<=n;++i)for(j=h[i];j;j=e[j].nx)++f[i][s[e[j].t]-'A'];
	for(i=1;i<=n;++i)if(!f[i][0]||!f[i][1])q[++qn]=i;
	for(i=1;i<=qn;++i)for(j=h[q[i]];j;j=e[j].nx)
		if(f[e[j].t][0]&&f[e[j].t][1]&&!--f[e[j].t][s[q[i]]-'A'])q[++qn]=e[j].t;
	puts(qn<n?"Yes":"No");
}
```