#include<bits/stdc++.h>
#define F first
#define S second
#define int long long
#define double long double
using namespace std;
typedef pair<int,int> P;
typedef pair<int,P> iP;
typedef pair<P,int> Pi;
typedef pair<P,P> PP;
double eps=1e-8;
int mod=1e9+7;
int inf=1LL<<55;

signed main(){
  int n,m;
  string s;
  vector<int> e[200000],v;
  cin>>n>>m>>s;
  for(int i=0,a,b;i<m;i++){
    cin>>a>>b;
    a--,b--;
    e[a].push_back(b);
    e[b].push_back(a);
  }

  queue<int> Q;
  int a[200000]={},b[200000]={},c=n;
  bool used[200000]={};

  for(int i=0;i<n;i++){
    for(int j=0;j<e[i].size();j++){
      if(s[e[i][j]]=='A')a[i]++;
      else b[i]++;
    }
    if(!a[i]||!b[i])Q.push(i),used[i]=1;
  }
  while(!Q.empty()){
    int p=Q.front();
    Q.pop();
    c--;
    for(int i=0;i<e[p].size();i++){
      int np=e[p][i];
      if(used[np])continue;
      if(s[p]=='A')a[np]--;
      else b[np]--;
      if(!a[np]||!b[np])Q.push(np),used[np]=1;
    }
  }
  cout<<(c?"Yes":"No")<<endl;
  return 0;
}

### 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>
#define F first
#define S second
#define int long long
#define double long double
using namespace std;
typedef pair<int,int> P;
typedef pair<int,P> iP;
typedef pair<P,int> Pi;
typedef pair<P,P> PP;
double eps=1e-8;
int mod=1e9+7;
int inf=1LL<<55;

signed main(){
  int n,m;
  string s;
  vector<int> e[200000],v;
  cin>>n>>m>>s;
  for(int i=0,a,b;i<m;i++){
    cin>>a>>b;
    a--,b--;
    e[a].push_back(b);
    e[b].push_back(a);
  }

  queue<int> Q;
  int a[200000]={},b[200000]={},c=n;
  bool used[200000]={};

  for(int i=0;i<n;i++){
    for(int j=0;j<e[i].size();j++){
      if(s[e[i][j]]=='A')a[i]++;
      else b[i]++;
    }
    if(!a[i]||!b[i])Q.push(i),used[i]=1;
  }
  while(!Q.empty()){
    int p=Q.front();
    Q.pop();
    c--;
    for(int i=0;i<e[p].size();i++){
      int np=e[p][i];
      if(used[np])continue;
      if(s[p]=='A')a[np]--;
      else b[np]--;
      if(!a[np]||!b[np])Q.push(np),used[np]=1;
    }
  }
  cout<<(c?"Yes":"No")<<endl;
  return 0;
}

```