#include<bits/stdc++.h>
#define rep(i,x,y) for (int i=(x);i<=(y);i++)
using namespace std;
const int N=2e5+5;
int n,m,x,y,rd[N][2],cnt,head[N],vis[N],num; char s[N];
struct edge{int to,nxt;}e[N<<1];
void adde(int x,int y){e[++cnt].to=y; e[cnt].nxt=head[x]; head[x]=cnt;}
int main(){
	scanf("%d%d%s",&n,&m,s+1);
	while (m--) scanf("%d%d",&x,&y),adde(x,y),adde(y,x),rd[x][s[y]-'A']++,rd[y][s[x]-'A']++;
	queue<int> q;
	rep (i,1,n) if (!rd[i][0]||!rd[i][1]) q.push(i),vis[i]=1;//只有连向B的边/A 
	while (!q.empty()){
		int u=q.front(); q.pop(); num++;
		for (int i=head[u],v;i;i=e[i].nxt){
			v=e[i].to;
			if (!--rd[v][s[u]-'A']&&!vis[v]) vis[v]=1,q.push(v);
		}
	}
	puts(num!=n?"Yes":"No");
	return 0;
}
//找形如AABB的环 
//用类似拓扑排序的方法做，对某个点记录A/B连向它的度 
### 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>
#define rep(i,x,y) for (int i=(x);i<=(y);i++)
using namespace std;
const int N=2e5+5;
int n,m,x,y,rd[N][2],cnt,head[N],vis[N],num; char s[N];
struct edge{int to,nxt;}e[N<<1];
void adde(int x,int y){e[++cnt].to=y; e[cnt].nxt=head[x]; head[x]=cnt;}
int main(){
	scanf("%d%d%s",&n,&m,s+1);
	while (m--) scanf("%d%d",&x,&y),adde(x,y),adde(y,x),rd[x][s[y]-'A']++,rd[y][s[x]-'A']++;
	queue<int> q;
	rep (i,1,n) if (!rd[i][0]||!rd[i][1]) q.push(i),vis[i]=1;//只有连向B的边/A 
	while (!q.empty()){
		int u=q.front(); q.pop(); num++;
		for (int i=head[u],v;i;i=e[i].nxt){
			v=e[i].to;
			if (!--rd[v][s[u]-'A']&&!vis[v]) vis[v]=1,q.push(v);
		}
	}
	puts(num!=n?"Yes":"No");
	return 0;
}
//找形如AABB的环 
//用类似拓扑排序的方法做，对某个点记录A/B连向它的度 
```