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.*;
  import java.util.*;
  public class Ishu
  {
      static Scanner scan = new Scanner(System.in);
      static BufferedWriter output = new BufferedWriter(new OutputStreamWriter(System.out)); 
      static void tc() throws Exception
      {
      int n = scan.nextInt();
      int k = scan.nextInt();
      long x = scan.nextLong();
      String str = scan.next();
      char[] ch = str.toCharArray();
      int i, j;
      
      List<Integer> list = new ArrayList<Integer>();
      int size = 0;
      for(i=0;i<n;++i)
          {
          if(ch[i] == 'a')
              {
              if(size > 0)
                  list.add(size);
              size = 0;
              continue;
              }
              
          ++size;     
          }
      if(size > 0)
          list.add(size);
          
      int m = list.size();
      int[] a = new int[m];
      for(i=0;i<m;++i)
          a[i] = list.get(i);
          
      long[] cnt = new long[m];    
      --x;
      for(i=m-1;i>=0;--i)    
          {
          int quo = a[i] * k + 1;
          long rem = x % quo;
          x -= rem;
          x /= (a[i] * k + 1);
          
          cnt[i] = rem;
          }
          
      j = 0;
      for(i=0;i<n;++i)
          {
          if(ch[i] == 'a')
              {
              output.write("a");
              continue;
              }
              
          while(cnt[j]-- > 0)
              output.write("b");    
              
          j++;
          while(i < n && ch[i] == '*')
              ++i;
              
          --i;
          }
       
      output.write("\n");   
      output.flush();
      }
      
      public static void main(String[] args) throws Exception
      {
      int t = 1;
      t = scan.nextInt();
      while(t-- > 0)
          tc();
      }
  }
  