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.*;
  import java.io.*;
  public class Main {
     // For fast input output
     static class FastReader {
         BufferedReader br;
         StringTokenizer st;
         public FastReader()
         { try {br = new BufferedReader(
         new FileReader("input.txt"));
         PrintStream out = new PrintStream(new FileOutputStream("output.txt"));
         System.setOut(out);}
         catch(Exception e) { 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;
         }
     }
      // end of fast i/o code
      public static void main(String[] args) {
          FastReader reader = new FastReader();
          StringBuilder sb = new StringBuilder("");
          
          int t = reader.nextInt();
          while(t-->0){
              int n = reader.nextInt();
              int k = reader.nextInt();
              long x = reader.nextLong();
              String s = reader.nextLine();
  
              StringBuilder ans = new StringBuilder("");
  
              x -= 1;
              for(int i=n-1; i>=0; i--){
                  if(s.charAt(i)=='a'){
                      ans.append('a');
                  }else{
                      int j = i;
                      long len = 0;
                      while(j>=0 && s.charAt(j)=='*'){
                          j--;
                          len++;
                      }
                      i = j + 1;
                      len = len*k + 1;
                      // System.out.println(t+" "+len+" "+x);
                      for(int it=1; it<=x%len; it++){
                          ans.append("b");
                      }
                      x = x/len;
                  }
              }
              ans.reverse();
              sb.append(ans+"\n");
          }
          System.out.println(sb);
      }
  }