#include <bits/stdc++.h>
using namespace std;
const int MOD=1e9+7;
//const int MOD=998244353;
const int INF=1e9;
const long long LINF=1e18;
#define int long long
//template
template <typename T>
void fin(T a){
  cout<<a<<endl;
  exit(0);
}
//main
signed main(){
  int N,M;cin>>N>>M;
  string s;cin>>s;
  std::vector<int> edge[N],r(N,0),b(N,0);
  while(M--){
    int u,v;cin>>u>>v;u--;v--;
    edge[u].push_back(v);
    edge[v].push_back(u);
    if(s[u]=='A')r[v]++;else b[v]++;
    if(s[v]=='A')r[u]++;else b[u]++;
  }
  queue<int> que;
  std::vector<int> dele(N,0);
  for(int i=0;i<N;i++)if(r[i]*b[i]==0)que.push(i);
  while(que.size()){
    int p=que.front();que.pop();
    if(dele[p]++)continue;
    for(int q:edge[p]){
      if(dele[q])continue;
      if(s[p]=='A')r[q]--;else b[q]--;
      if(r[q]*b[q]==0)que.push(q);
    }
  }
  for(int i=0;i<N;i++)if(!dele[i])fin("Yes");
  fin("No");
}

### Prompt

In cpp, your task is to solve 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;
const int MOD=1e9+7;
//const int MOD=998244353;
const int INF=1e9;
const long long LINF=1e18;
#define int long long
//template
template <typename T>
void fin(T a){
  cout<<a<<endl;
  exit(0);
}
//main
signed main(){
  int N,M;cin>>N>>M;
  string s;cin>>s;
  std::vector<int> edge[N],r(N,0),b(N,0);
  while(M--){
    int u,v;cin>>u>>v;u--;v--;
    edge[u].push_back(v);
    edge[v].push_back(u);
    if(s[u]=='A')r[v]++;else b[v]++;
    if(s[v]=='A')r[u]++;else b[u]++;
  }
  queue<int> que;
  std::vector<int> dele(N,0);
  for(int i=0;i<N;i++)if(r[i]*b[i]==0)que.push(i);
  while(que.size()){
    int p=que.front();que.pop();
    if(dele[p]++)continue;
    for(int q:edge[p]){
      if(dele[q])continue;
      if(s[p]=='A')r[q]--;else b[q]--;
      if(r[q]*b[q]==0)que.push(q);
    }
  }
  for(int i=0;i<N;i++)if(!dele[i])fin("Yes");
  fin("No");
}

```