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: //BA-String
  import java.io.*;
  import java.util.*;
  
  public class C1620{
     
     public static void main(String[] args)throws IOException{
        BufferedReader f = new BufferedReader(new InputStreamReader(System.in));
        PrintWriter out = new PrintWriter(System.out);
        
        int t = Integer.parseInt(f.readLine());
        
        for(int q = 1; q <= t; q++){
        
           StringTokenizer st = new StringTokenizer(f.readLine());
        
           int n = Integer.parseInt(st.nextToken());
           long m = Long.parseLong(st.nextToken());
           long x = Long.parseLong(st.nextToken());
           
           String s = f.readLine();
           char[] array = s.toCharArray();
           
           ArrayList<Long> alist = new ArrayList<Long>();
           ArrayList<Character> alist2 = new ArrayList<Character>();                //collapse * to 1
           for(int k = 0; k < n; k++){
              if(array[k] == '*'){
                 if(k == 0 || array[k-1] != '*'){
                    alist.add(1L);
                    alist2.add('*');
                 }
                 else alist.set(alist.size()-1,alist.get(alist.size()-1)+1L);
              } else {
                 alist2.add('a');
              }
           }
           
           if(alist.size() == 0){
              out.println(s);
              continue;
           }
           
           if(m == 0){
              StringJoiner sj = new StringJoiner("");
              for(int k = 0; k < n; k++) if(array[k] == 'a') sj.add("a");
              out.println(sj.toString());
              continue;
           }
           
           long[] muls = new long[alist.size()];
           Arrays.fill(muls,Long.MAX_VALUE);
           muls[alist.size()-1] = 1L;
           long multiplier = alist.get(alist.size()-1)*m+1;
           for(int k = alist.size()-2; k >= 0; k--){
              muls[k] = multiplier;
              multiplier *= alist.get(k)*m+1;
              if(multiplier > x || multiplier < 0){
                 break;
              }
           }
           
           long[] bs = new long[alist.size()];
           
           for(int k = 0; k < alist.size()-1; k++){
              if(muls[k] > x) continue;
              bs[k] = (x-1)/muls[k];
              x-=bs[k]*muls[k];
           }
           bs[alist.size()-1] = x-1;
           
           int bi = 0;
           
           StringJoiner sj = new StringJoiner("");
           for(int k = 0; k < alist2.size(); k++){
              if(alist2.get(k) == 'a'){
                 sj.add("a");
              } else {
                 for(int j = 0; j < bs[bi]; j++){
                    sj.add("b");
                 }
                 bi++;
              }
           }
           
           out.println(sj.toString());
              
           
           
           
        }
        
        
        
        
        out.close();
     }
     
        
  }