#include <bits/stdc++.h>
using namespace std;
const int N=200005;
struct Gragh{
	static const int M=N*2;
	int cnt,y[M],nxt[M],fst[N];
	void add(int a,int b){
		y[++cnt]=b,nxt[cnt]=fst[a],fst[a]=cnt;
	}
}g;
int n,m,c[N],out[N][2],del[N];
char s[N];
int q[N],head,tail;
int main(){
	scanf("%d%d",&n,&m);
	scanf("%s",s+1);
	for (int i=1;i<=n;i++)
		if (s[i]=='A')
			c[i]=0;
		else
			c[i]=1;
	memset(out,0,sizeof out);
	for (int i=1;i<=m;i++){
		int a,b;
		scanf("%d%d",&a,&b);
		out[a][c[b]]++;
		out[b][c[a]]++;
		g.add(a,b);
		g.add(b,a);
	}
	memset(del,0,sizeof del);
	head=tail=0;
	for (int i=1;i<=n;i++)
		if (!out[i][0]||!out[i][1])
			q[++tail]=i,del[i]=1;
	while (head<tail){
		int x=q[++head];
		for (int i=g.fst[x];i;i=g.nxt[i]){
			int y=g.y[i];
			if (del[y])
				continue;
			out[y][c[x]]--;
			if (!out[y][c[x]])
				q[++tail]=y,del[y]=1;
		}
	}
	puts(tail!=n?"Yes":"No");
	return 0;
}
### Prompt

Please provide a CPP coded solution 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;
const int N=200005;
struct Gragh{
	static const int M=N*2;
	int cnt,y[M],nxt[M],fst[N];
	void add(int a,int b){
		y[++cnt]=b,nxt[cnt]=fst[a],fst[a]=cnt;
	}
}g;
int n,m,c[N],out[N][2],del[N];
char s[N];
int q[N],head,tail;
int main(){
	scanf("%d%d",&n,&m);
	scanf("%s",s+1);
	for (int i=1;i<=n;i++)
		if (s[i]=='A')
			c[i]=0;
		else
			c[i]=1;
	memset(out,0,sizeof out);
	for (int i=1;i<=m;i++){
		int a,b;
		scanf("%d%d",&a,&b);
		out[a][c[b]]++;
		out[b][c[a]]++;
		g.add(a,b);
		g.add(b,a);
	}
	memset(del,0,sizeof del);
	head=tail=0;
	for (int i=1;i<=n;i++)
		if (!out[i][0]||!out[i][1])
			q[++tail]=i,del[i]=1;
	while (head<tail){
		int x=q[++head];
		for (int i=g.fst[x];i;i=g.nxt[i]){
			int y=g.y[i];
			if (del[y])
				continue;
			out[y][c[x]]--;
			if (!out[y][c[x]])
				q[++tail]=y,del[y]=1;
		}
	}
	puts(tail!=n?"Yes":"No");
	return 0;
}
```