#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef vector<ll> vec;
typedef vector<vec> mat;
typedef pair<ll,ll> pll;
const ll mod=1e9+7;
//const ll mod=998244353;
const ll inf=1LL<<61;

int main() {
  ll n,m;
  string s;
  cin >> n >> m >> s;
  s="#"+s;
  mat v(n+1);
  mat deg(n+1,vec(2));
  vec q,vis(n+1);
  for(ll i=m,x,y;i--;) {
    cin >> x >> y;
    v[x].push_back(y);
    v[y].push_back(x);
    deg[x][s[y]-'A']++;
    deg[y][s[x]-'A']++;
  }
  for(ll i=1;i<=n;i++) {
    if(!deg[i][0]||!deg[i][1]) {
      q.push_back(i);
      vis[i]=1;
    }
  }
  for(ll i=0;i<q.size();i++) {
    for(auto j:v[q[i]]) {
      if(!--deg[j][s[q[i]]-'A']&&!vis[j]) {
        q.push_back(j);
        vis[j]=1;
      }
    }
  }
  cout << (q.size()<n?"Yes":"No") << endl;
}
### Prompt

Your task is to create 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;
typedef long long ll;
typedef vector<ll> vec;
typedef vector<vec> mat;
typedef pair<ll,ll> pll;
const ll mod=1e9+7;
//const ll mod=998244353;
const ll inf=1LL<<61;

int main() {
  ll n,m;
  string s;
  cin >> n >> m >> s;
  s="#"+s;
  mat v(n+1);
  mat deg(n+1,vec(2));
  vec q,vis(n+1);
  for(ll i=m,x,y;i--;) {
    cin >> x >> y;
    v[x].push_back(y);
    v[y].push_back(x);
    deg[x][s[y]-'A']++;
    deg[y][s[x]-'A']++;
  }
  for(ll i=1;i<=n;i++) {
    if(!deg[i][0]||!deg[i][1]) {
      q.push_back(i);
      vis[i]=1;
    }
  }
  for(ll i=0;i<q.size();i++) {
    for(auto j:v[q[i]]) {
      if(!--deg[j][s[q[i]]-'A']&&!vis[j]) {
        q.push_back(j);
        vis[j]=1;
      }
    }
  }
  cout << (q.size()<n?"Yes":"No") << endl;
}
```