#include<iostream>
#include<vector>
#include<queue>
using namespace std;
const int N=2e5+2;
int cnt[N][2];
bool check[N];
vector<int> adj[N];
queue<int> lis;
int main(){
  ios::sync_with_stdio(0);
  cin.tie(0);
  int n,i,j,k,l,m;
  string s;
  cin>>n>>m>>s;
  s=" "+s;
  for(i=1;i<=m;i++){
    cin>>j>>k;
    adj[j].push_back(k);
    adj[k].push_back(j);
    cnt[k][(int)(s[j]-'A')]++;
    cnt[j][(int)(s[k]-'A')]++;
  }
  for(i=1;i<=n;i++){
    if(!cnt[i][0]||!cnt[i][1]){
      check[i]=true;
      lis.push(i);
    }
  }
  while(lis.size()){
    l=lis.front();
    lis.pop();
    for(i=0;i<adj[l].size();i++){
      j=adj[l][i];
      cnt[j][(int)(s[l]-'A')]--;
      if(!cnt[j][(int)(s[l]-'A')]&&!check[j]){
        check[j]=true;
        lis.push(j);
      }
    }
  }
  for(i=1;i<=n;i++){
    if(!check[i]){
      cout<<"Yes";
        return 0;
    }
  }
  cout<<"No";
}
### Prompt

Please create a solution in Cpp to 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<iostream>
#include<vector>
#include<queue>
using namespace std;
const int N=2e5+2;
int cnt[N][2];
bool check[N];
vector<int> adj[N];
queue<int> lis;
int main(){
  ios::sync_with_stdio(0);
  cin.tie(0);
  int n,i,j,k,l,m;
  string s;
  cin>>n>>m>>s;
  s=" "+s;
  for(i=1;i<=m;i++){
    cin>>j>>k;
    adj[j].push_back(k);
    adj[k].push_back(j);
    cnt[k][(int)(s[j]-'A')]++;
    cnt[j][(int)(s[k]-'A')]++;
  }
  for(i=1;i<=n;i++){
    if(!cnt[i][0]||!cnt[i][1]){
      check[i]=true;
      lis.push(i);
    }
  }
  while(lis.size()){
    l=lis.front();
    lis.pop();
    for(i=0;i<adj[l].size();i++){
      j=adj[l][i];
      cnt[j][(int)(s[l]-'A')]--;
      if(!cnt[j][(int)(s[l]-'A')]&&!check[j]){
        check[j]=true;
        lis.push(j);
      }
    }
  }
  for(i=1;i<=n;i++){
    if(!check[i]){
      cout<<"Yes";
        return 0;
    }
  }
  cout<<"No";
}
```