#include <bits/stdc++.h>
#define FOR(i, a, b) for(int i=(int)(a); i!=(int)(b); i++)
#define RFOR(i, a, b) for(int i=(int)(a); i!=(int)(b); i--)
#define REP(i, n) FOR(i, 0, n)
using namespace std;
using ll = long long;

constexpr int lim = 2*1e5;
int a[lim+1], b[lim+1];
int main(){
  int n, m; cin >> n >> m;
  string s; cin >> s;

  set<int> g[n];

  REP(i, m){
    int x, y; cin >> x >> y;
    x--; y--;
    g[x].insert(y);
    g[y].insert(x);
  }

  set<int> q;
  REP(i, n){
    for(auto x: g[i]){
      if(s[x] == 'A') a[i]++;
      else b[i]++;
    }
    if(a[i] == 0 || b[i] == 0) q.insert(i);
  }
  
  int cnt = 0;
  while(q.size()){
    int v = *q.begin(); 
    q.erase(q.begin());
    cnt++;
    for(auto x: g[v]){
      if(x == v) continue;

      g[x].erase(v);
      if(s[v] == 'A') a[x]--;
      else b[x]--;

      if(a[x] == 0 || b[x] == 0) q.insert(x);
    }
  }

  if(cnt == n) cout << "No" << endl;
  else cout << "Yes" << endl;

  return 0;
}

### Prompt

Generate 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>
#define FOR(i, a, b) for(int i=(int)(a); i!=(int)(b); i++)
#define RFOR(i, a, b) for(int i=(int)(a); i!=(int)(b); i--)
#define REP(i, n) FOR(i, 0, n)
using namespace std;
using ll = long long;

constexpr int lim = 2*1e5;
int a[lim+1], b[lim+1];
int main(){
  int n, m; cin >> n >> m;
  string s; cin >> s;

  set<int> g[n];

  REP(i, m){
    int x, y; cin >> x >> y;
    x--; y--;
    g[x].insert(y);
    g[y].insert(x);
  }

  set<int> q;
  REP(i, n){
    for(auto x: g[i]){
      if(s[x] == 'A') a[i]++;
      else b[i]++;
    }
    if(a[i] == 0 || b[i] == 0) q.insert(i);
  }
  
  int cnt = 0;
  while(q.size()){
    int v = *q.begin(); 
    q.erase(q.begin());
    cnt++;
    for(auto x: g[v]){
      if(x == v) continue;

      g[x].erase(v);
      if(s[v] == 'A') a[x]--;
      else b[x]--;

      if(a[x] == 0 || b[x] == 0) q.insert(x);
    }
  }

  if(cnt == n) cout << "No" << endl;
  else cout << "Yes" << endl;

  return 0;
}

```