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.*;
  
  public class Main {
      public static BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
      public static BufferedWriter bw = new BufferedWriter(new OutputStreamWriter(System.out));
      public static String[] input;
      public static int n;
      public static int k;
      public static long x;
      public static String str;
      public static void main(String[] args) throws IOException {
          int t = Integer.parseInt(br.readLine());
          while(t-- > 0) {
              input = br.readLine().split(" ");
              n = Integer.parseInt(input[0]);
              k = Integer.parseInt(input[1]);
              x = Long.parseLong(input[2])-1;
              str = br.readLine();
              int len = str.length();
              char prev = str.charAt(len-1);
              int cnt = 0;
              StringBuilder ans = new StringBuilder();
              if(prev == '*')
                  ++cnt;
              else
                  ans.append('a');
          
              for(int i = len-2; i >= 0; --i) {
                  if(prev != '*' && str.charAt(i) == '*') {
                      cnt = 1;
                      prev = '*';
                  }
                  else if(prev == '*' && str.charAt(i) != '*') {
                      int num = cnt*k + 1;
                      long b = x % num;
                      while(b-- > 0)
                          ans.append('b');
                      x /= num;
                      prev = 'a';
                      cnt = 0;
                      ans.append('a');
                  }
                  else if(prev == '*' && str.charAt(i) == '*')
                      ++cnt;
                  else
                      ans.append('a');
              }
              if(cnt > 0) {
                  int num = cnt*k + 1;
                  long b = x % num;
                  while(b-- > 0)
                      ans.append('b');
              }
              
              bw.write(ans.reverse().toString() + "\n");
          }
          br.close(); 
          bw.close();
          return;
      }
  }