#include<cstdio>
#include<queue>
#include<vector>
#include<algorithm>
using namespace std;
int n,m,x,y,i,j,col[200005],aa[2][200005],cnt=0,vis[200005],h,v;
vector<int> a[200005];
queue<int> q;
char ss[200005];
int main()
{
	scanf("%d%d",&n,&m);
	scanf("%s",ss);
	for(i=1;i<=n;i++)col[i]=ss[i-1]-'A';
	for(i=1;i<=m;i++)
	{
		scanf("%d%d",&x,&y);
		if(x==y)aa[col[x]][x]++;
		else
		{
			aa[col[x]][y]++;
			aa[col[y]][x]++;
			a[x].push_back(y);
			a[y].push_back(x);
		}
	}
	for(i=1;i<=n;i++)
	{
		if(!aa[0][i]||!aa[1][i])
		{
			q.push(i);
			cnt++;
			vis[i]=1;
		}
	}
	while(!q.empty())
	{
		h=q.front();
		q.pop();
		for(i=0;i<a[h].size();i++)
		{
			v=a[h][i];
			if(!vis[v])
			{
				aa[col[h]][v]--;
				if(!aa[0][v]||!aa[1][v])
				{
				q.push(v);
				cnt++;
				vis[v]=1;
				}
			}
		}
	}
	if(cnt>=n)
	{
		printf("No\n");
		
	}
	else printf("Yes\n");
	return 0;
}
### Prompt

In cpp, your task is to solve 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<cstdio>
#include<queue>
#include<vector>
#include<algorithm>
using namespace std;
int n,m,x,y,i,j,col[200005],aa[2][200005],cnt=0,vis[200005],h,v;
vector<int> a[200005];
queue<int> q;
char ss[200005];
int main()
{
	scanf("%d%d",&n,&m);
	scanf("%s",ss);
	for(i=1;i<=n;i++)col[i]=ss[i-1]-'A';
	for(i=1;i<=m;i++)
	{
		scanf("%d%d",&x,&y);
		if(x==y)aa[col[x]][x]++;
		else
		{
			aa[col[x]][y]++;
			aa[col[y]][x]++;
			a[x].push_back(y);
			a[y].push_back(x);
		}
	}
	for(i=1;i<=n;i++)
	{
		if(!aa[0][i]||!aa[1][i])
		{
			q.push(i);
			cnt++;
			vis[i]=1;
		}
	}
	while(!q.empty())
	{
		h=q.front();
		q.pop();
		for(i=0;i<a[h].size();i++)
		{
			v=a[h][i];
			if(!vis[v])
			{
				aa[col[h]][v]--;
				if(!aa[0][v]||!aa[1][v])
				{
				q.push(v);
				cnt++;
				vis[v]=1;
				}
			}
		}
	}
	if(cnt>=n)
	{
		printf("No\n");
		
	}
	else printf("Yes\n");
	return 0;
}
```