description: You are given an integer k and a string s that consists only of characters 'a' (a lowercase Latin letter) and '*' (an asterisk).
  
  Each asterisk should be replaced with several (from 0 to k inclusive) lowercase Latin letters 'b'. Different asterisk can be replaced with different counts of letter 'b'.
  
  The result of the replacement is called a BA-string.
  
  Two strings a and b are different if they either have different lengths or there exists such a position i that a_i ≠ b_i.
  
  A string a is lexicographically smaller than a string b if and only if one of the following holds: 
  
    * a is a prefix of b, but a ≠ b; 
    * in the first position where a and b differ, the string a has a letter that appears earlier in the alphabet than the corresponding letter in b. 
  
  
  
  Now consider all different BA-strings and find the x-th lexicographically smallest of them.
  
  Input
  
  The first line contains a single integer t (1 ≤ t ≤ 2000) — the number of testcases.
  
  The first line of each testcase contains three integers n, k and x (1 ≤ n ≤ 2000; 0 ≤ k ≤ 2000; 1 ≤ x ≤ 10^{18}). n is the length of string s.
  
  The second line of each testcase is a string s. It consists of n characters, each of them is either 'a' (a lowercase Latin letter) or '*' (an asterisk).
  
  The sum of n over all testcases doesn't exceed 2000. For each testcase x doesn't exceed the total number of different BA-strings. String s contains at least one character 'a'.
  
  Output
  
  For each testcase, print a single string, consisting only of characters 'b' and 'a' (lowercase Latin letters) — the x-th lexicographically smallest BA-string.
  
  Example
  
  Input
  
  
  3
  2 4 3
  a*
  4 1 3
  a**a
  6 3 20
  **a***
  
  
  Output
  
  
  abb
  abba
  babbbbbbbbb
  
  Note
  
  In the first testcase of the example, BA-strings ordered lexicographically are: 
  
    1. a
    2. ab
    3. abb
    4. abbb
    5. abbbb
  
  
  
  In the second testcase of the example, BA-strings ordered lexicographically are: 
  
    1. aa
    2. aba
    3. abba
  
  
  
  Note that string "aba" is only counted once, even though there are two ways to replace asterisks with characters 'b' to get it.
solution: 
  
  import java.util.Arrays;
  import java.util.LinkedList;
  import java.util.Scanner;
  
  public class C {
      public static void main(String[] args) {
          Scanner s = new Scanner(System.in);
          int lines = s.nextInt();
  
          s.nextLine();
  
          for (int i = 0; i < lines; i += 1) {
              solve(s.nextInt(), s.nextInt(), s.nextLong(), s.next());
          }
      }
  
  
      public static void solve(int len, int k, long x, String s) {
          x--;
          LinkedList<Integer> count = new LinkedList<>();
  
          int c = 0;
          for (int i = 0; i < s.length(); i++) {
              if (s.charAt(i) == '*') c++;
              else {
                  if (c == 0) continue;
                  count.add(c);
                  c = 0;
              }
          }
          if (c != 0) count.add(c);
  
          int[] num = new int[count.size()];
          for (int i = num.length - 1; i >= 0; i--) {
              num[i] = (int) (x % (k * count.get(i) + 1));
              x /= (k * count.get(i) + 1);
          }
  
          StringBuilder erg = new StringBuilder();
          c = 0;
          for (int i = 0; i < s.length(); i++) {
              if (s.charAt(i) == 'a') erg.append('a'); else erg.append("b".repeat(num[c++]));
              while (i <= s.length() - 2) if (s.charAt(i + 1) == '*' && s.charAt(i) == '*') i++; else break;
          }
  
          System.out.println(erg);
      }
  }
  