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.BufferedReader;
  import java.io.IOException;
  import java.io.InputStreamReader;
  import java.io.PrintWriter;
  import java.util.ArrayList;
  import java.util.*;
  
  public class BAString {
          public static void main(String[] args) throws IOException {
              BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
              int t = Integer.parseInt(br.readLine().split(" ")[0]);
              PrintWriter pr = new PrintWriter(System.out);
              while (t > 0) {
                  t--;
                  String[] ints = br.readLine().split(" ");
                  Long n = Long.valueOf(ints[0]);
                  Long k = Long.valueOf(ints[1]);
                  Long x = Long.valueOf(ints[2]);
                  String s = br.readLine();
                  int count = 0;
                  List<Integer> l = new ArrayList<>();
                  for(int i = 0;i<s.length();i++){
                      if (s.charAt(i) == '*'){
                          count++;
                      }
                      else{
                          if (count > 0){
                              l.add(count);
                              count = 0;
                          }
                      }
                  }
                  //to cover trailing *
                  x--;
                  if (count >0) l.add(count);
                  List<Long> digits = new ArrayList<>();
                  for(int i = l.size()-1;i>=0;i--){
                      long range = l.get(i)*k + 1;
                      long digit = x%(range);
                      x = x / range;
                      digits.add(0,digit);
                  }
                  StringBuffer sb = new StringBuffer();
                  int digCount = 0;
                  for(int i = 0;i<s.length();i++){
                      if (s.charAt(i) == 'a'){
                          sb.append('a');
                      }
                      else{
                          for(int j =0;j<digits.get(digCount);j++){
                              sb.append('b');
                          }
                          digCount++;
                          while(i < s.length() && s.charAt(i) == '*'){
                              i++;
                          }
                          i--;
                      }
                  }
                  for(int i = 0;i<sb.length();i++){
                      pr.print(sb.charAt(i));
                  }
                  pr.println();
              }
              pr.close();
          }
  
  }
  