#include<bits/stdc++.h>
using namespace std;
using ll = long long;

vector<int> g[212345];

int main() {
  std::ios::sync_with_stdio(false), std::cin.tie(0);
  int n, m; cin>>n>>m;
  string s;
  cin >> 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);
  }
  vector<int> ok(n, 1);
  // vector<int> used(n);
  // int col = 1;
  function<void(int)> check = [&](int i) {
    if(!ok[i]) return;
    int f[2] = {};
    for(int j : g[i]) {
      if(ok[j]) f[s[j] == 'A']++;
    }
    if(!(f[0] && f[1])) ok[i] = 0;
    if(!ok[i]) {
      auto newg = g[i];
      for(int j : g[i]) {
        if(!ok[j]) continue;
        check(j);
        if(ok[j]) newg.push_back(j);
      }
      g[i] = newg;
    }
  };
  for(int i = 0; i < n; i++) check(i);
  cout << (find((ok).begin(), (ok).end(), 1) != ok.end() ? "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>
using namespace std;
using ll = long long;

vector<int> g[212345];

int main() {
  std::ios::sync_with_stdio(false), std::cin.tie(0);
  int n, m; cin>>n>>m;
  string s;
  cin >> 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);
  }
  vector<int> ok(n, 1);
  // vector<int> used(n);
  // int col = 1;
  function<void(int)> check = [&](int i) {
    if(!ok[i]) return;
    int f[2] = {};
    for(int j : g[i]) {
      if(ok[j]) f[s[j] == 'A']++;
    }
    if(!(f[0] && f[1])) ok[i] = 0;
    if(!ok[i]) {
      auto newg = g[i];
      for(int j : g[i]) {
        if(!ok[j]) continue;
        check(j);
        if(ok[j]) newg.push_back(j);
      }
      g[i] = newg;
    }
  };
  for(int i = 0; i < n; i++) check(i);
  cout << (find((ok).begin(), (ok).end(), 1) != ok.end() ? "Yes" : "No") << endl;
  return 0;
}

```