# include<bits/stdc++.h>
using namespace std;
const int MAX=4e5+5;
struct p{
	int x,y;
}c[MAX<<1];
int n,m,num;
int h[MAX],vis[MAX];
char s[MAX];
bool use[MAX];
void add(int x=0,int y=0)
{
	scanf("%d%d",&x,&y);
	if(s[x]==s[y]) c[++num]=(p){h[x],y+n},h[x]=num,c[++num]=(p){h[y],x+n},h[y]=num;
	else c[++num]=(p){h[x+n],y},h[x+n]=num,c[++num]=(p){h[y+n],x},h[y+n]=num;
}
void dfs(int x)
{
	vis[x]=1,use[x]=1;
	for(int i=h[x];i;i=c[i].x)
	  if(!vis[c[i].y]) dfs(c[i].y);
	  else if(use[c[i].y]) printf("Yes"),exit(0);
	use[x]=0;
}
int main()
{
	scanf("%d%d%s",&n,&m,s+1);
	for(int i=1;i<=m;add(),++i);
	for(int i=1;i<=n;++i)
	  if(!vis[i]) dfs(i);
	return printf("No"),0;
}
### 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;
const int MAX=4e5+5;
struct p{
	int x,y;
}c[MAX<<1];
int n,m,num;
int h[MAX],vis[MAX];
char s[MAX];
bool use[MAX];
void add(int x=0,int y=0)
{
	scanf("%d%d",&x,&y);
	if(s[x]==s[y]) c[++num]=(p){h[x],y+n},h[x]=num,c[++num]=(p){h[y],x+n},h[y]=num;
	else c[++num]=(p){h[x+n],y},h[x+n]=num,c[++num]=(p){h[y+n],x},h[y+n]=num;
}
void dfs(int x)
{
	vis[x]=1,use[x]=1;
	for(int i=h[x];i;i=c[i].x)
	  if(!vis[c[i].y]) dfs(c[i].y);
	  else if(use[c[i].y]) printf("Yes"),exit(0);
	use[x]=0;
}
int main()
{
	scanf("%d%d%s",&n,&m,s+1);
	for(int i=1;i<=m;add(),++i);
	for(int i=1;i<=n;++i)
	  if(!vis[i]) dfs(i);
	return printf("No"),0;
}
```