#include<bits/stdc++.h>
using namespace std;
const int N=2e5+10;
struct edge{
	int v,nxt;
}g[N<<1];
int tot=0,head[N];
inline void add(int u,int v){
	g[++tot]=edge{v,head[u]}; head[u]=tot;
	g[++tot]=edge{u,head[v]}; head[v]=tot;
}
int n,m,dgr[N][2],q[N],h,t,vis[N];
char s[N];
int main(){
	scanf("%d%d%s",&n,&m,s);
	for(int i=1; i<=m; ++i){
		int u,v; scanf("%d%d",&u,&v);
		add(u,v); dgr[u][s[v-1]=='B']++; dgr[v][s[u-1]=='B']++;
	}
	h=t=0; for(int i=1; i<=n; ++i)if(!dgr[i][0]||!dgr[i][1])q[++t]=i,vis[i]=1;
	while(h<t){
		int u=q[++h];
		for(int i=head[u]; i; i=g[i].nxt){
			int v=g[i].v; if(vis[v])continue;
			--dgr[v][s[u-1]=='B'];
			if(!dgr[v][s[u-1]=='B']){
				vis[v]=1; q[++t]=v;
			}
		}
	}
	puts(t==n?"No":"Yes");
	return 0;
}
### Prompt

Create a solution in Cpp for 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 N=2e5+10;
struct edge{
	int v,nxt;
}g[N<<1];
int tot=0,head[N];
inline void add(int u,int v){
	g[++tot]=edge{v,head[u]}; head[u]=tot;
	g[++tot]=edge{u,head[v]}; head[v]=tot;
}
int n,m,dgr[N][2],q[N],h,t,vis[N];
char s[N];
int main(){
	scanf("%d%d%s",&n,&m,s);
	for(int i=1; i<=m; ++i){
		int u,v; scanf("%d%d",&u,&v);
		add(u,v); dgr[u][s[v-1]=='B']++; dgr[v][s[u-1]=='B']++;
	}
	h=t=0; for(int i=1; i<=n; ++i)if(!dgr[i][0]||!dgr[i][1])q[++t]=i,vis[i]=1;
	while(h<t){
		int u=q[++h];
		for(int i=head[u]; i; i=g[i].nxt){
			int v=g[i].v; if(vis[v])continue;
			--dgr[v][s[u-1]=='B'];
			if(!dgr[v][s[u-1]=='B']){
				vis[v]=1; q[++t]=v;
			}
		}
	}
	puts(t==n?"No":"Yes");
	return 0;
}
```