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

const int N = 2e5+10;
vector< vector<int> > G(N);
vector< vector<int> > c(N,vector<int>(2));
//c[u][0]=隣接するAの数,c[u][1]=隣接するBの数}
vector<bool> deleted(N);
string s;

void check(int i){
  if(deleted[i]) return;
  deleted[i] = true;
  for(int v : G[i]){
    if(deleted[v]) continue;
    (s[i] == 'A' ? c[v][0] : c[v][1]) -= 1;
    if(c[v][0] == 0 || c[v][1] == 0) check(v);
  }
}

signed main(){
  cin.tie(0);
  ios::sync_with_stdio(false);

  int n,m; cin >> n >> m >> s;
  for(int i = 0; i < m; i++){
    int a,b; cin >> a >> b;
    a--,b--;
    G[a].push_back(b);
    G[b].push_back(a);
    (s[a] == 'A' ? c[b][0] : c[b][1]) += 1;
    (s[b] == 'A' ? c[a][0] : c[a][1]) += 1;
  }

  for(int i = 0; i < n; i++){
    if(c[i][0] == 0 || c[i][1] == 0) check(i);
  }

  for(int i = 0; i < n; i++){
    if(!deleted[i]){
      cout << "Yes" << endl;
      return 0;
    }
  }
  cout << "No" << 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;
#define int long long

const int N = 2e5+10;
vector< vector<int> > G(N);
vector< vector<int> > c(N,vector<int>(2));
//c[u][0]=隣接するAの数,c[u][1]=隣接するBの数}
vector<bool> deleted(N);
string s;

void check(int i){
  if(deleted[i]) return;
  deleted[i] = true;
  for(int v : G[i]){
    if(deleted[v]) continue;
    (s[i] == 'A' ? c[v][0] : c[v][1]) -= 1;
    if(c[v][0] == 0 || c[v][1] == 0) check(v);
  }
}

signed main(){
  cin.tie(0);
  ios::sync_with_stdio(false);

  int n,m; cin >> n >> m >> s;
  for(int i = 0; i < m; i++){
    int a,b; cin >> a >> b;
    a--,b--;
    G[a].push_back(b);
    G[b].push_back(a);
    (s[a] == 'A' ? c[b][0] : c[b][1]) += 1;
    (s[b] == 'A' ? c[a][0] : c[a][1]) += 1;
  }

  for(int i = 0; i < n; i++){
    if(c[i][0] == 0 || c[i][1] == 0) check(i);
  }

  for(int i = 0; i < n; i++){
    if(!deleted[i]){
      cout << "Yes" << endl;
      return 0;
    }
  }
  cout << "No" << endl;

  return 0;
}

```