#include<iostream>
#include<cstdio>
#include<queue>
#include<cstring>
#define M 500010
using namespace std;

int n,m,num,cnt;
int head[M],c[M],in[M][2];
char s[M];bool vis[M];
struct point{int to,next;}e[M<<1];
void add(int from,int to)
{
	e[++num].next=head[from];
	e[num].to=to;
	head[from]=num;
}
queue<int>q;
int main()
{
	scanf("%d%d",&n,&m);
	scanf("%s",s+1);
	for(int i=1;i<=n;i++) c[i]=s[i]-'A';
	for(int i=1;i<=m;i++)
	{
		int x,y;scanf("%d%d",&x,&y);
		add(x,y); add(y,x);
		in[x][c[y]]++; in[y][c[x]]++;
	}
	for(int i=1;i<=n;i++)
		if(!in[i][0]||!in[i][1])
		{
			q.push(i);
			cnt++;
			vis[i]=true;
		}
	while(!q.empty())
	{
		int x=q.front();q.pop();
		for(int i=head[x];i;i=e[i].next)
		{
			int to=e[i].to;
			in[to][c[x]]--;
			if((!in[to][0]||!in[to][1])&&!vis[to])
			{
				q.push(to);
				cnt++;
				vis[to]=true;
			}
		}
	}
	printf(cnt==n?"No":"Yes");
	return 0;
}
### Prompt

Develop a solution in cpp to the problem described below:

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<queue>
#include<cstring>
#define M 500010
using namespace std;

int n,m,num,cnt;
int head[M],c[M],in[M][2];
char s[M];bool vis[M];
struct point{int to,next;}e[M<<1];
void add(int from,int to)
{
	e[++num].next=head[from];
	e[num].to=to;
	head[from]=num;
}
queue<int>q;
int main()
{
	scanf("%d%d",&n,&m);
	scanf("%s",s+1);
	for(int i=1;i<=n;i++) c[i]=s[i]-'A';
	for(int i=1;i<=m;i++)
	{
		int x,y;scanf("%d%d",&x,&y);
		add(x,y); add(y,x);
		in[x][c[y]]++; in[y][c[x]]++;
	}
	for(int i=1;i<=n;i++)
		if(!in[i][0]||!in[i][1])
		{
			q.push(i);
			cnt++;
			vis[i]=true;
		}
	while(!q.empty())
	{
		int x=q.front();q.pop();
		for(int i=head[x];i;i=e[i].next)
		{
			int to=e[i].to;
			in[to][c[x]]--;
			if((!in[to][0]||!in[to][1])&&!vis[to])
			{
				q.push(to);
				cnt++;
				vis[to]=true;
			}
		}
	}
	printf(cnt==n?"No":"Yes");
	return 0;
}
```