#include <bits/stdc++.h>
using namespace std;
#define N 200100

struct edge {int nex,to;}e[N<<1];
int n,m,cnt,he,ta;
char s[N];
int sum[N][2],head[N],q[N],inq[N];

void add(int x,int y) {e[++cnt].to=y; e[cnt].nex=head[x]; head[x]=cnt;}

int x,y;
int main()
{
	scanf("%d%d",&n,&m);
	scanf("%s",s+1);
	for (int i=1;i<=m;i++) {
		scanf("%d%d",&x,&y);
		sum[x][s[y]-'A']++;
		sum[y][s[x]-'A']++;
		add(x,y); add(y,x);
	}
	for (int i=1;i<=n;i++)
		if (!sum[i][0] || !sum[i][1]) {
			q[++ta]=i;
			inq[i]=1;
		}
	
	while (he<ta)
	{
		int x=q[++he];
		for (int i=head[x];i;i=e[i].nex)
		{
			int v=e[i].to;
			if (!inq[v])
				if (--sum[v][s[x]-'A']==0) 
					q[++ta]=v,inq[v]=1;
		}
	}
	
	for (int i=1;i<=n;i++)
		if (!inq[i])
			return puts("Yes"),0;
	return puts("No"),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;
#define N 200100

struct edge {int nex,to;}e[N<<1];
int n,m,cnt,he,ta;
char s[N];
int sum[N][2],head[N],q[N],inq[N];

void add(int x,int y) {e[++cnt].to=y; e[cnt].nex=head[x]; head[x]=cnt;}

int x,y;
int main()
{
	scanf("%d%d",&n,&m);
	scanf("%s",s+1);
	for (int i=1;i<=m;i++) {
		scanf("%d%d",&x,&y);
		sum[x][s[y]-'A']++;
		sum[y][s[x]-'A']++;
		add(x,y); add(y,x);
	}
	for (int i=1;i<=n;i++)
		if (!sum[i][0] || !sum[i][1]) {
			q[++ta]=i;
			inq[i]=1;
		}
	
	while (he<ta)
	{
		int x=q[++he];
		for (int i=head[x];i;i=e[i].nex)
		{
			int v=e[i].to;
			if (!inq[v])
				if (--sum[v][s[x]-'A']==0) 
					q[++ta]=v,inq[v]=1;
		}
	}
	
	for (int i=1;i<=n;i++)
		if (!inq[i])
			return puts("Yes"),0;
	return puts("No"),0;
}
```