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

int main() {
   ios_base::sync_with_stdio(false);
   int n, m; cin >> n >> m;
   string s; cin >> s;
   vector<int> a(n);
   for (int i = 0; i < n; ++i) a[i] = s[i] - 'A';
   vector<vector<int>> g(n);
   vector<vector<int>> deg(n, vector<int>(2, 0));
   for (int i = 0; i < m; ++i) {
      int u, v; cin >> u >> v; u--, v--;
      deg[u][a[v]]++;
      deg[v][a[u]]++;
      g[u].push_back(v);
      g[v].push_back(u);
   }
   vector<int> was(n, 0);
   vector<int> q;
   for (int i = 0; i < n; ++i) {
      if (deg[i][0] == 0 || deg[i][1] == 0) {
         was[i] = 1;
         q.push_back(i);
      }
   }
   for (int i = 0; i < q.size(); ++i) {
      int u = q[i];
      for (int v : g[u]) {
         deg[v][a[u]]--;
         if (deg[v][0] == 0 || deg[v][1] == 0) {
            if (was[v] == 0) {
               was[v] = 1;
               q.push_back(v);
            }
         }
      }
   }
   cout << (q.size() == n ? "No" : "Yes") << '\n';
}

### Prompt

Please formulate 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;

int main() {
   ios_base::sync_with_stdio(false);
   int n, m; cin >> n >> m;
   string s; cin >> s;
   vector<int> a(n);
   for (int i = 0; i < n; ++i) a[i] = s[i] - 'A';
   vector<vector<int>> g(n);
   vector<vector<int>> deg(n, vector<int>(2, 0));
   for (int i = 0; i < m; ++i) {
      int u, v; cin >> u >> v; u--, v--;
      deg[u][a[v]]++;
      deg[v][a[u]]++;
      g[u].push_back(v);
      g[v].push_back(u);
   }
   vector<int> was(n, 0);
   vector<int> q;
   for (int i = 0; i < n; ++i) {
      if (deg[i][0] == 0 || deg[i][1] == 0) {
         was[i] = 1;
         q.push_back(i);
      }
   }
   for (int i = 0; i < q.size(); ++i) {
      int u = q[i];
      for (int v : g[u]) {
         deg[v][a[u]]--;
         if (deg[v][0] == 0 || deg[v][1] == 0) {
            if (was[v] == 0) {
               was[v] = 1;
               q.push_back(v);
            }
         }
      }
   }
   cout << (q.size() == n ? "No" : "Yes") << '\n';
}

```