#include <iostream>
#include <algorithm>
#include <string>
#include <vector>
#include <math.h>
#include <queue>
#define MOD 1000000007
typedef long long ll;
using namespace std;
const int N=2e5+1;

int n,m;
int a[N],b[N];
string s;
vector<int> G[N];
bool no_ab[N];

int main(){
  cin>>n>>m;
  cin>>s;

  for(int i=0;i<m;i++){
    int c,d;
    cin>>c>>d;
    c--;
    d--;
    G[c].push_back(d);
    G[d].push_back(c);
    if(s[c]=='A') a[d]++;
    else b[d]++;
    if(s[d]=='A') a[c]++;
    else b[c]++;
  }

  queue<int> Q;
  for(int i=0;i<n;i++){
    if(a[i]==0||b[i]==0){
      no_ab[i]=1;
      Q.push(i);
    }
  }

  while(!Q.empty()){
    int u=Q.front();
    Q.pop();
    for(int v:G[u]){
      if(!no_ab[v]){
        if(s[u]=='A') a[v]--;
        else b[v]--;
        if(a[v]==0||b[v]==0){
          no_ab[v]=1;
          Q.push(v);
        }
      }
    }
  }

  bool flag=0;
  for(int i=0;i<n;i++){
    if(!no_ab[i]) flag=1;
  }

  if(flag) cout<<"Yes"<<endl;
  else cout<<"No"<<endl;

  return 0;
}

### 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 <algorithm>
#include <string>
#include <vector>
#include <math.h>
#include <queue>
#define MOD 1000000007
typedef long long ll;
using namespace std;
const int N=2e5+1;

int n,m;
int a[N],b[N];
string s;
vector<int> G[N];
bool no_ab[N];

int main(){
  cin>>n>>m;
  cin>>s;

  for(int i=0;i<m;i++){
    int c,d;
    cin>>c>>d;
    c--;
    d--;
    G[c].push_back(d);
    G[d].push_back(c);
    if(s[c]=='A') a[d]++;
    else b[d]++;
    if(s[d]=='A') a[c]++;
    else b[c]++;
  }

  queue<int> Q;
  for(int i=0;i<n;i++){
    if(a[i]==0||b[i]==0){
      no_ab[i]=1;
      Q.push(i);
    }
  }

  while(!Q.empty()){
    int u=Q.front();
    Q.pop();
    for(int v:G[u]){
      if(!no_ab[v]){
        if(s[u]=='A') a[v]--;
        else b[v]--;
        if(a[v]==0||b[v]==0){
          no_ab[v]=1;
          Q.push(v);
        }
      }
    }
  }

  bool flag=0;
  for(int i=0;i<n;i++){
    if(!no_ab[i]) flag=1;
  }

  if(flag) cout<<"Yes"<<endl;
  else cout<<"No"<<endl;

  return 0;
}

```