#include <bits/stdc++.h>
using namespace std;
const int maxn=200005*2;
int n,m,vis[maxn];
vector<int>G[maxn];
char str[maxn];
void dfs(int x){
  vis[x]=1;
  for(int v:G[x])if(~vis[v]){
    if(vis[v]==1){
      puts("Yes");
      exit(0);
    }else{
      dfs(v);
    }
  }
  vis[x]=-1;
}
int main(){
  //freopen("aa.in","r",stdin);
  scanf("%d%d%s",&n,&m,str+1);
  for(int i=1;i<=m;i++){
    int x,y;scanf("%d%d",&x,&y);
    // printf("e %d %d\n",x,y);
    if(str[x]==str[y]){
      G[x].push_back(n+y);
      G[y].push_back(n+x);
    }else{
      G[n+x].push_back(y);
      G[n+y].push_back(x);
    }
  }
  for(int i=1;i<=n;i++)if(!vis[i])dfs(i);
  puts("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 maxn=200005*2;
int n,m,vis[maxn];
vector<int>G[maxn];
char str[maxn];
void dfs(int x){
  vis[x]=1;
  for(int v:G[x])if(~vis[v]){
    if(vis[v]==1){
      puts("Yes");
      exit(0);
    }else{
      dfs(v);
    }
  }
  vis[x]=-1;
}
int main(){
  //freopen("aa.in","r",stdin);
  scanf("%d%d%s",&n,&m,str+1);
  for(int i=1;i<=m;i++){
    int x,y;scanf("%d%d",&x,&y);
    // printf("e %d %d\n",x,y);
    if(str[x]==str[y]){
      G[x].push_back(n+y);
      G[y].push_back(n+x);
    }else{
      G[n+x].push_back(y);
      G[n+y].push_back(x);
    }
  }
  for(int i=1;i<=n;i++)if(!vis[i])dfs(i);
  puts("No");
  return 0;
}

```