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.io.*;
  import java.util.*;
  
  public class Main {
      public static void main(String[] args) {
          FastIO fio = new FastIO();
  
          int t = fio.nextInt();
  
          for (int i = 0; i < t; i++) {
              int n = fio.nextInt();
              int k = fio.nextInt();
              long x = fio.nextLong() - 1;
              String s = fio.nextLine();
  
              StringBuilder reverse = new StringBuilder();
  
              for (int j = s.length() - 1; j >= 0; j--) {
                  char c = s.charAt(j);
                  if (c == '*') {
                      int length = 0;
  
                      while (j >= 0 && s.charAt(j) == '*') {
                          j--;
                          length++;
                      }
  
                      j++; // to offset the j-- in the for-loop
                      int div = (length * k) + 1;
                      long mod = x % div;
  
                      for (int l = 0; l < mod; l++) {
                          reverse.append("b");
                      }
  
                      x /= div;
                  } else {
                      reverse.append(c);
                  }
              }
  
              StringBuilder result = new StringBuilder();
  
              for (int j = reverse.length() - 1; j >= 0; j--) {
                  result.append(reverse.charAt(j));
              }
  
              fio.println(result);
          }
  
          fio.close();
      }
  }
  
  /**
   * Fast I/O
   * @source https://www.geeksforgeeks.org/fast-io-in-java-in-competitive-programming/
   */
  class FastIO extends PrintWriter 
  { 
      BufferedReader br; 
      StringTokenizer st;
  
      public FastIO() 
      { 
          super(new BufferedOutputStream(System.out)); 
          br = new BufferedReader(new
                  InputStreamReader(System.in));
      } 
  
      String next() 
      { 
          while (st == null || !st.hasMoreElements()) 
          { 
              try
              { 
                  st = new StringTokenizer(br.readLine()); 
              } 
              catch (IOException  e) 
              { 
                  e.printStackTrace(); 
              } 
          } 
          return st.nextToken(); 
      } 
  
      int nextInt() 
      { 
          return Integer.parseInt(next()); 
      } 
  
      long nextLong() 
      { 
          return Long.parseLong(next()); 
      } 
  
      double nextDouble() 
      { 
          return Double.parseDouble(next()); 
      } 
  
      String nextLine() 
      { 
          String str = ""; 
          try
          { 
              str = br.readLine(); 
          } 
          catch (IOException e) 
          { 
              e.printStackTrace(); 
          } 
          return str; 
      } 
  }
  
  