#include<bits/stdc++.h>
using namespace std;
using Int = long long;
//INSERT ABOVE HERE
signed main(){
  Int n,m;
  cin>>n>>m;
  string s;
  cin>>s;
  vector<Int> A(n,0),B(n,0);
  vector<set<Int> > G(n);
  for(Int i=0;i<m;i++){
    Int x,y;
    cin>>x>>y;
    x--;y--;
    if(G[x].count(y)) continue;
    G[x].emplace(y);
    G[y].emplace(x);
    if(s[x]=='A') A[y]++;
    if(s[x]=='B') B[y]++;
    if(s[y]=='A') A[x]++;
    if(s[y]=='B') B[x]++;
  }
  vector<Int> used(n,0);
  queue<Int> q;
  for(Int i=0;i<n;i++){
    if(A[i]&&B[i]) continue;
    used[i]=1;
    q.emplace(i);
  }
  while(!q.empty()){
    Int v=q.front();q.pop();
    for(Int u:G[v]){
      if(used[u]) continue;
      if(s[v]=='A') A[u]--;
      if(s[v]=='B') B[u]--;
      G[u].erase(v);
      if(A[u]&&B[u]) continue;
      used[u]=1;
      q.emplace(u);
    }
  }
  Int ans=0;
  for(Int i=0;i<n;i++) ans|=!used[i];
  cout<<(ans?"Yes":"No")<<endl;
  return 0;
}

### Prompt

Your task is to create 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;
using Int = long long;
//INSERT ABOVE HERE
signed main(){
  Int n,m;
  cin>>n>>m;
  string s;
  cin>>s;
  vector<Int> A(n,0),B(n,0);
  vector<set<Int> > G(n);
  for(Int i=0;i<m;i++){
    Int x,y;
    cin>>x>>y;
    x--;y--;
    if(G[x].count(y)) continue;
    G[x].emplace(y);
    G[y].emplace(x);
    if(s[x]=='A') A[y]++;
    if(s[x]=='B') B[y]++;
    if(s[y]=='A') A[x]++;
    if(s[y]=='B') B[x]++;
  }
  vector<Int> used(n,0);
  queue<Int> q;
  for(Int i=0;i<n;i++){
    if(A[i]&&B[i]) continue;
    used[i]=1;
    q.emplace(i);
  }
  while(!q.empty()){
    Int v=q.front();q.pop();
    for(Int u:G[v]){
      if(used[u]) continue;
      if(s[v]=='A') A[u]--;
      if(s[v]=='B') B[u]--;
      G[u].erase(v);
      if(A[u]&&B[u]) continue;
      used[u]=1;
      q.emplace(u);
    }
  }
  Int ans=0;
  for(Int i=0;i<n;i++) ans|=!used[i];
  cout<<(ans?"Yes":"No")<<endl;
  return 0;
}

```