#include<bits/stdc++.h>
using namespace std;

int N,M;
string s;
int ac[200005];
int bc[200005];
vector<int> G[200005];

int main(){
  cin>>N>>M;
  cin>>s;
  for(int i=0;i<M;i++){
    int u,v;
    cin>>u>>v;
    u--,v--;
    G[u].push_back(v);
    G[v].push_back(u);

    if(s[v]=='A')ac[u]++;
    else bc[u]++;
    
    if(s[u]=='A')ac[v]++;
    else bc[v]++;
  }
  
  set<int> st;
  queue< int > Q;
  
  for(int i=0;i<N;i++){
    if( ac[i]==0 || bc[i]==0){
      Q.push(i);
      st.insert(i);
    }
  }
  while(!Q.empty()){
    int pos=Q.front();Q.pop();
    for(int to:G[pos]){
      if(s[pos]=='A')ac[to]--;
      else bc[to]--;
      if(ac[to]==0||bc[to]==0)
        if( st.count(to) == 0 ){
          Q.push(to);
          st.insert(to);
        }
    }
  }

  if(st.size() == N )cout<<"No"<<endl;
  else cout<<"Yes"<<endl;
  return 0;
}

### Prompt

Your challenge is to write a cpp solution 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<bits/stdc++.h>
using namespace std;

int N,M;
string s;
int ac[200005];
int bc[200005];
vector<int> G[200005];

int main(){
  cin>>N>>M;
  cin>>s;
  for(int i=0;i<M;i++){
    int u,v;
    cin>>u>>v;
    u--,v--;
    G[u].push_back(v);
    G[v].push_back(u);

    if(s[v]=='A')ac[u]++;
    else bc[u]++;
    
    if(s[u]=='A')ac[v]++;
    else bc[v]++;
  }
  
  set<int> st;
  queue< int > Q;
  
  for(int i=0;i<N;i++){
    if( ac[i]==0 || bc[i]==0){
      Q.push(i);
      st.insert(i);
    }
  }
  while(!Q.empty()){
    int pos=Q.front();Q.pop();
    for(int to:G[pos]){
      if(s[pos]=='A')ac[to]--;
      else bc[to]--;
      if(ac[to]==0||bc[to]==0)
        if( st.count(to) == 0 ){
          Q.push(to);
          st.insert(to);
        }
    }
  }

  if(st.size() == N )cout<<"No"<<endl;
  else cout<<"Yes"<<endl;
  return 0;
}

```