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: /*
    Challenge 1: Newbie to CM in 1year (Dec 2021 - Nov 2022)   🔥 5* Codechef
    Challenge 2: CM to IM in 1 year (Dec 2022 - Nov 2023)    🔥🔥 6* Codechef
    Challenge 3: IM to GM in 1 year (Dec 2023 - Nov 2024)  🔥🔥🔥 7* Codechef
    Goal: Become better in CP!
    Key: Consistency!
  */
  
  import java.util.*;
  import java.io.*;
  import java.math.*;
  
  public class Coder {
  
    static long x;
    static int n, k;
    static char c[];
    static StringBuffer str=new StringBuffer();
  
    static void solve(){
      StringBuffer ans=new StringBuffer();
      x--;
      for(int i=n-1;i>=0;i--){
        if(c[i]=='a') ans.append('a');
        else{
          int len=0;
          while(i>=0 && c[i]=='*'){len++;i--;}
          long rem=x%(len*k+1);
          x/=(len*k+1);
          i++;
          for(int k=0;k<rem;k++) ans.append('b');
        }
      }
      str.append(ans.reverse()).append("\n");
    }
  
    public static void main(String[] args) throws java.lang.Exception {
      BufferedReader bf;
      PrintWriter pw;
      boolean lenv=false;
      if(lenv){
        bf = new BufferedReader(
                            new FileReader("input.txt"));
        pw=new PrintWriter(new
              BufferedWriter(new FileWriter("output.txt")));
      }else{
        bf = new BufferedReader(new InputStreamReader(System.in));
        pw = new PrintWriter(new OutputStreamWriter(System.out));
      }
      
      int q = Integer.parseInt(bf.readLine().trim());
      while (q-- > 0) {
        String st[]=bf.readLine().trim().split("\\s+");
        n=Integer.parseInt(st[0]);
        k=Integer.parseInt(st[1]);
        x=Long.parseLong(st[2]);
        c=bf.readLine().trim().toCharArray();
        solve();
      }
      pw.println(str);
      pw.flush();
      // System.outin.print(str);
    }
  }