#include <bits/stdc++.h>

using namespace std;

int main(int argc, char *argv[]) {
  std::ios_base::sync_with_stdio(false);
  int n, m;
  string s;
  cin >> n >> m >> s;
  vector<int> a(n);
  for (int i = 0; i < n; i++) {
    a[i] = s[i] == 'A';
  }
  vector<vector<int>> b(n, vector<int>(2));
  vector<vector<int>> g(n);
  for (int i = 0; i < m; i++) {
    int u, v;
    cin >> u >> v;
    u--, v--;
    g[u].push_back(v);
    g[v].push_back(u);
    b[u][a[v]]++;
    b[v][a[u]]++;
  }
  vector<bool> vis(n);
  queue<int> q;
  for (int i = 0; i < n; i++) {
    if (!b[i][0] || !b[i][1]) {
      q.push(i);
    }
  }
  while (!q.empty()) {
    int u = q.front();
    q.pop();
    vis[u] = true;
    for (auto v : g[u]) {
      if (!vis[v]) {
        b[v][a[u]]--;
        if (b[v][a[u]] == 0) {
          q.push(v);
        }
      }
    }
  }
  bool ok = true;
  for (auto x : vis) {
    ok &= x;
  }
  cout << (!ok ? "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;

int main(int argc, char *argv[]) {
  std::ios_base::sync_with_stdio(false);
  int n, m;
  string s;
  cin >> n >> m >> s;
  vector<int> a(n);
  for (int i = 0; i < n; i++) {
    a[i] = s[i] == 'A';
  }
  vector<vector<int>> b(n, vector<int>(2));
  vector<vector<int>> g(n);
  for (int i = 0; i < m; i++) {
    int u, v;
    cin >> u >> v;
    u--, v--;
    g[u].push_back(v);
    g[v].push_back(u);
    b[u][a[v]]++;
    b[v][a[u]]++;
  }
  vector<bool> vis(n);
  queue<int> q;
  for (int i = 0; i < n; i++) {
    if (!b[i][0] || !b[i][1]) {
      q.push(i);
    }
  }
  while (!q.empty()) {
    int u = q.front();
    q.pop();
    vis[u] = true;
    for (auto v : g[u]) {
      if (!vis[v]) {
        b[v][a[u]]--;
        if (b[v][a[u]] == 0) {
          q.push(v);
        }
      }
    }
  }
  bool ok = true;
  for (auto x : vis) {
    ok &= x;
  }
  cout << (!ok ? "Yes" : "No") << endl; 
  return 0;
}

```