#include<cstdio>
#include<algorithm>
#include<iostream>
#include<cstring>
#include<vector>
#include<queue>
using namespace std;
typedef long long LL;
const int N=200005;
int n,m;
char ss[N];
int X[N],Y[N];
int du[N][2];
vector<int> vec[N];
queue<int> q;
bool in[N];
int main()
{
	scanf("%d%d",&n,&m);
	scanf("%s",ss+1);
	for (int u=1;u<=m;u++)
	{
		int x,y;
		scanf("%d%d",&x,&y);
		vec[x].push_back(y);
		vec[y].push_back(x);
		du[y][ss[x]-'A']++;
		du[x][ss[y]-'A']++;
	}
	memset(in,true,sizeof(in));
	for (int u=1;u<=n;u++)
	{
		if (du[u][0]<=0||du[u][1]<=0)
		{
			in[u]=false;
			q.push(u);
		}
	}
	while (!q.empty())
	{
		int x=q.front();q.pop();
		int siz=vec[x].size();
		for (int u=0;u<siz;u++)
		{
			int y=vec[x][u];
			if (in[y]==false) continue;
			du[y][ss[x]-'A']--;
			if (du[y][ss[x]-'A']==0)	{in[y]=false;q.push(y);}
		}
	}
	for (int u=1;u<=n;u++) if (in[u])
	{
		printf("Yes\n");
		return 0;
	}
	printf("No\n");
	return 0;
}
### Prompt

Please create a solution in Cpp to 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<algorithm>
#include<iostream>
#include<cstring>
#include<vector>
#include<queue>
using namespace std;
typedef long long LL;
const int N=200005;
int n,m;
char ss[N];
int X[N],Y[N];
int du[N][2];
vector<int> vec[N];
queue<int> q;
bool in[N];
int main()
{
	scanf("%d%d",&n,&m);
	scanf("%s",ss+1);
	for (int u=1;u<=m;u++)
	{
		int x,y;
		scanf("%d%d",&x,&y);
		vec[x].push_back(y);
		vec[y].push_back(x);
		du[y][ss[x]-'A']++;
		du[x][ss[y]-'A']++;
	}
	memset(in,true,sizeof(in));
	for (int u=1;u<=n;u++)
	{
		if (du[u][0]<=0||du[u][1]<=0)
		{
			in[u]=false;
			q.push(u);
		}
	}
	while (!q.empty())
	{
		int x=q.front();q.pop();
		int siz=vec[x].size();
		for (int u=0;u<siz;u++)
		{
			int y=vec[x][u];
			if (in[y]==false) continue;
			du[y][ss[x]-'A']--;
			if (du[y][ss[x]-'A']==0)	{in[y]=false;q.push(y);}
		}
	}
	for (int u=1;u<=n;u++) if (in[u])
	{
		printf("Yes\n");
		return 0;
	}
	printf("No\n");
	return 0;
}
```