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.util.ArrayList;
          import java.util.Arrays;
          import java.util.StringTokenizer;
  
  public class bastring {
      public static void main(String[] args) throws IOException {
          BufferedReader f = new BufferedReader(new InputStreamReader(System.in));
          int tcase = Integer.parseInt(f.readLine());
          StringBuilder out = new StringBuilder();
          for(int p=0; p<tcase; p++){
              StringTokenizer st = new StringTokenizer(f.readLine());
              int len = Integer.parseInt(st.nextToken());
              int k = Integer.parseInt(st.nextToken());
              long x = Long.parseLong(st.nextToken());
              String s = f.readLine();
  
              String[] eek =s.split("a");
  
              ArrayList<Long> gaps = new ArrayList<>();
              for(int i=0; i<eek.length; i++){
                  if(eek[i].length()!=0){
  
                      gaps.add((long) eek[i].length() * k + 1);
  
                  }
              }
              long[] ans = new long[gaps.size()];
  
              long num = 1;
              int ind = gaps.size()-1;
              while (ind>-1&&x/gaps.get(ind)>=num){
                  long j=num;
                  num*=gaps.get(ind);
                  ans[ind]=j;
                  ind--;
              }
  
              x--;
              if(ind>-1)
                  ans[ind]=num;
  
              for(int i=0; i<gaps.size(); i++){
                  if(ans[i]!=0){
                      long y = x/ans[i];
                      x%=ans[i];
                      ans[i]=y;
  
                  }
              }
              StringBuilder sb = new StringBuilder();
              char[] str = s.toCharArray();
              int ansind = 0;
              for(int i=0; i<str.length; i++){
                  if(str[i]=='a'){
                      sb.append('a');
                  }else{
                      if(i==0||str[i-1]=='a'){
                          for(int j=0; j<ans[ansind]; j++){
                              sb.append('b');
                          }
                          ansind++;
                      }
                  }
              }
              out.append(sb+"\n");
  
          }
          System.out.println(out);
      }
  }
  