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.util.*;
  
  public class Main {
      public static void main(String[] args) throws Exception {
  
          final Scanner cin = new Scanner(System.in);
          long t = cin.nextLong();
          for (long ii = 0; ii < t; ii++) {
              StringBuilder cout = new StringBuilder();
              long n = cin.nextLong();
              long k = cin.nextLong();
              long x = cin.nextLong();
              x--;
              cin.nextLine();
              String str = cin.nextLine();
              int i;
              long ctr = 0, mul = 1;
              ArrayList<Long> arr = new ArrayList<>();
              for (i = 0; i < str.length(); i++) {
                  if (str.charAt(i) == 'a') {
                      if (ctr > 0)
                          arr.add(ctr);
                      ctr = 0;
                      arr.add(-1L);
                  } else
                      ctr++;
              }
              if (ctr > 0)
                  arr.add(ctr);
              for (i = arr.size() - 1; i >= 0; i--) {
                  if (arr.get(i) != -1L) {
                      ctr = x % (1 + k * arr.get(i));
                      for (; ctr > 0; ctr--)
                          cout.append('b');
                      x = x / (1 + k * arr.get(i));
                  } else {
                      cout.append('a');
                  }
              }
              System.out.println(cout.reverse());
          }
      }
  }
  