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.*;
  import java.io.*;
  
  public class _1620_C {
      public static void main(String[] args) throws IOException {
          BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
          PrintWriter out = new PrintWriter(System.out);
          int t = Integer.parseInt(in.readLine());
          while(t-- > 0) {
              StringTokenizer line = new StringTokenizer(in.readLine());
              int n = Integer.parseInt(line.nextToken());
              int k = Integer.parseInt(line.nextToken());
              long x = Long.parseLong(line.nextToken());
              String s = in.readLine();
              ArrayList<Integer> digits = new ArrayList<Integer>();
              for(int i = 0; i < n; i++) {
                  if(s.charAt(i) == '*') {
                      if(i > 0 && s.charAt(i - 1) == '*') {
                          int ind = digits.size() - 1;
                          digits.set(ind, digits.get(ind) + k);
                      }else {
                          digits.add(k);
                      }
                  }
              }
              if(digits.size() == 0) {
                  out.println(s);
                  continue;
              }
              long[] num = new long[digits.size()];
              num[digits.size() - 1] = x - 1;
              for(int i = digits.size() - 1; i > 0; i--) {
                  int cur = digits.get(i) + 1;
                  num[i - 1] += num[i] / cur;
                  num[i] %= cur;
              }
              StringBuilder sb = new StringBuilder();
              int p = 0;
              for(int i = 0; i < n; i++) {
                  if(s.charAt(i) == 'a') {
                      sb.append('a');
                  }else {
                      if(i == 0 || s.charAt(i - 1) == 'a') {
                          for(int j = 0; j < num[p]; j++) {
                              sb.append('b');
                          }
                          p++;
                      }
                  }
              }
              out.println(sb.toString());
          }
          in.close();
          out.close();
      }
  }
  