#include<bits/stdc++.h>
using namespace std;

const int maxn=200010;
struct Edge
{
	int v,next;
}	e[maxn<<1];
int g[maxn],tot;
void addedge(int x,int y)
{
	tot++;
	e[tot].v=y;
	e[tot].next=g[x];
	g[x]=tot;
}
int du[maxn][2];
char s[maxn];
int que[maxn],cnt;
bool vis[maxn];
int n,m;

int main()
{
	scanf("%d%d",&n,&m);
	scanf("%s",s+1);
	for (int i=1;i<=m;i++)
	{
		int x,y;
		scanf("%d%d",&x,&y);
		addedge(x,y);
		addedge(y,x);
		du[x][s[y]-'A']++;
		du[y][s[x]-'A']++;
	}
	for (int i=1;i<=n;i++)
	if (du[i][0]==0 || du[i][1]==0)
	{
		que[++cnt]=i;
		vis[i]=true;
	}
	for (int i=1;i<=cnt;i++)
	{
		int u=que[i];
		for (int j=g[u];j;j=e[j].next)
			if (!vis[e[j].v])
			{
				du[e[j].v][s[u]-'A']--;
				if (du[e[j].v][s[u]-'A']==0)
				{
					que[++cnt]=e[j].v;
					vis[e[j].v]=true;
				}
			}
	}
	for (int i=1;i<=n;i++)
		if (!vis[i])
		{
			printf("Yes\n");
			return 0;
		}
	printf("No\n");
	return 0;
}
### Prompt

Your challenge is to write a Cpp solution 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<bits/stdc++.h>
using namespace std;

const int maxn=200010;
struct Edge
{
	int v,next;
}	e[maxn<<1];
int g[maxn],tot;
void addedge(int x,int y)
{
	tot++;
	e[tot].v=y;
	e[tot].next=g[x];
	g[x]=tot;
}
int du[maxn][2];
char s[maxn];
int que[maxn],cnt;
bool vis[maxn];
int n,m;

int main()
{
	scanf("%d%d",&n,&m);
	scanf("%s",s+1);
	for (int i=1;i<=m;i++)
	{
		int x,y;
		scanf("%d%d",&x,&y);
		addedge(x,y);
		addedge(y,x);
		du[x][s[y]-'A']++;
		du[y][s[x]-'A']++;
	}
	for (int i=1;i<=n;i++)
	if (du[i][0]==0 || du[i][1]==0)
	{
		que[++cnt]=i;
		vis[i]=true;
	}
	for (int i=1;i<=cnt;i++)
	{
		int u=que[i];
		for (int j=g[u];j;j=e[j].next)
			if (!vis[e[j].v])
			{
				du[e[j].v][s[u]-'A']--;
				if (du[e[j].v][s[u]-'A']==0)
				{
					que[++cnt]=e[j].v;
					vis[e[j].v]=true;
				}
			}
	}
	for (int i=1;i<=n;i++)
		if (!vis[i])
		{
			printf("Yes\n");
			return 0;
		}
	printf("No\n");
	return 0;
}
```