#include <bits/stdc++.h>

using namespace std;

const long long linf = 1e18;

int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  
  int n, m;
  cin >> n >> m;
  string s;
  cin >> s;
  vector<vector<int>> g(n);
  
  for (int i = 0; i < m; ++i) {
    int x, y;
    cin >> x >> y;
    --x; --y;
    g[x].push_back(y);
    g[y].push_back(x);
  }
  
  queue<int> q;
  vector<int> a(n), b(n);
  vector<int> used(n);
  
  for (int v = 0; v < n; ++v) {
    for (int to: g[v]) {
      if (s[to] == 'A')
        ++a[v];
      else
        ++b[v];
    }
    if (!a[v] || !b[v]) {
      q.push(v);
      used[v] = 1;
    }
  }
  
  int cnt = 0;
  while (!q.empty()) {
    int v = q.front();
    q.pop();
    ++cnt;
    
    for (int to: g[v]) {
      if (used[to])
        continue;
      if (s[v] == 'A')
        --a[to];
      else
        --b[to];
      if (!a[to] || !b[to]) {
        q.push(to);
        used[to] = 1;
      }
    }
  }
  
  cout << (cnt < n ? "Yes" : "No") << endl;
}

### 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 <bits/stdc++.h>

using namespace std;

const long long linf = 1e18;

int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  
  int n, m;
  cin >> n >> m;
  string s;
  cin >> s;
  vector<vector<int>> g(n);
  
  for (int i = 0; i < m; ++i) {
    int x, y;
    cin >> x >> y;
    --x; --y;
    g[x].push_back(y);
    g[y].push_back(x);
  }
  
  queue<int> q;
  vector<int> a(n), b(n);
  vector<int> used(n);
  
  for (int v = 0; v < n; ++v) {
    for (int to: g[v]) {
      if (s[to] == 'A')
        ++a[v];
      else
        ++b[v];
    }
    if (!a[v] || !b[v]) {
      q.push(v);
      used[v] = 1;
    }
  }
  
  int cnt = 0;
  while (!q.empty()) {
    int v = q.front();
    q.pop();
    ++cnt;
    
    for (int to: g[v]) {
      if (used[to])
        continue;
      if (s[v] == 'A')
        --a[to];
      else
        --b[to];
      if (!a[to] || !b[to]) {
        q.push(to);
        used[to] = 1;
      }
    }
  }
  
  cout << (cnt < n ? "Yes" : "No") << endl;
}

```