#include<bits/stdc++.h>
using namespace std;
#define RI register int
int read() {
	int q=0;char ch=' ';
	while(ch<'0'||ch>'9') ch=getchar();
	while(ch>='0'&&ch<='9') q=q*10+ch-'0',ch=getchar();
	return q;
}
const int N=200005;
char s[N];int tot,top,n,m,js;
int st[N],h[N],ne[N<<1],to[N<<1],va[N],vb[N],ban[N];
void add(int x,int y) {to[++tot]=y,ne[tot]=h[x],h[x]=tot;}
int main()
{
	int x,y;
	n=read(),m=read();
	scanf("%s",s+1);
	for(RI i=1;i<=m;++i) {
		x=read(),y=read(),add(x,y),add(y,x);
		if(s[x]=='A') ++va[y]; else ++vb[y];
		if(s[y]=='A') ++va[x]; else ++vb[x];
	}
	for(RI i=1;i<=n;++i) if(!va[i]||!vb[i]) st[++top]=i,ban[i]=1;
	while(top) {
		x=st[top],++js,--top;
		for(RI i=h[x];i;i=ne[i]) {
			if(ban[to[i]]) continue;
			if(s[x]=='A') --va[to[i]];
			else --vb[to[i]];
			if(!va[to[i]]||!vb[to[i]]) st[++top]=to[i],ban[to[i]]=1;
		}
	}
	if(js<n) puts("Yes");
	else puts("No");
	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<bits/stdc++.h>
using namespace std;
#define RI register int
int read() {
	int q=0;char ch=' ';
	while(ch<'0'||ch>'9') ch=getchar();
	while(ch>='0'&&ch<='9') q=q*10+ch-'0',ch=getchar();
	return q;
}
const int N=200005;
char s[N];int tot,top,n,m,js;
int st[N],h[N],ne[N<<1],to[N<<1],va[N],vb[N],ban[N];
void add(int x,int y) {to[++tot]=y,ne[tot]=h[x],h[x]=tot;}
int main()
{
	int x,y;
	n=read(),m=read();
	scanf("%s",s+1);
	for(RI i=1;i<=m;++i) {
		x=read(),y=read(),add(x,y),add(y,x);
		if(s[x]=='A') ++va[y]; else ++vb[y];
		if(s[y]=='A') ++va[x]; else ++vb[x];
	}
	for(RI i=1;i<=n;++i) if(!va[i]||!vb[i]) st[++top]=i,ban[i]=1;
	while(top) {
		x=st[top],++js,--top;
		for(RI i=h[x];i;i=ne[i]) {
			if(ban[to[i]]) continue;
			if(s[x]=='A') --va[to[i]];
			else --vb[to[i]];
			if(!va[to[i]]||!vb[to[i]]) st[++top]=to[i],ban[to[i]]=1;
		}
	}
	if(js<n) puts("Yes");
	else puts("No");
	return 0;
}
```