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: //package com.company;
  
  import java.util.*;
  import java.io.BufferedReader;
  import java.io.IOException;
  import java.io.InputStream;
  import java.io.InputStreamReader;
  import java.util.StringTokenizer;
  
  public class C {
  
      static class scanner {
          static BufferedReader reader;
          static StringTokenizer tokenizer;
          static void init(InputStream input) {
              reader = new BufferedReader(new InputStreamReader(input));
              tokenizer = new StringTokenizer("");
          }
          static String next() throws IOException {
              while (!tokenizer.hasMoreTokens()) {
                  tokenizer = new StringTokenizer(reader.readLine());
              }
              return tokenizer.nextToken();
          }
          static int nextInt() throws IOException {return Integer.parseInt(next());}
          static double nextDouble() throws IOException {return Double.parseDouble(next());}
          static long nextLong() throws IOException {return Long.parseLong(next());}
      }
      static long min(long a, long b) {return Math.min(a, b);}
      static long min(long a, long b, long c) {return min(a, min(b, c));}
      static int min(int a, int b) {return Math.min(a, b);}
      static int min(int a, int b, int c) {return min(a, min(b, c));}
      static long max(long a, long b) {return Math.max(a, b);}
      static long max(long a, long b, long c) {return max(a, max(b, c));}
      static int max(int a, int b) {return Math.max(a, b);}
      static int max(int a, int b, int c) {return max(a, max(b, c));}
      static int abs(int x) {return Math.abs(x);}
      static long abs(long x) {return Math.abs(x);}
      static long ceil(double x) {return (long) Math.ceil(x);}
      static long floor(double x) {return (long) Math.floor(x);}
      static double sqrt(int x) {return Math.sqrt(x);}
      static int gcd(int a, int b) {if(b == 0)return a;return gcd(b, a % b);}
      static long gcd(long a, long b) {if(b == 0)return a;return gcd(b, a % b);}
  
  
  
      public static void main(String[] args) throws IOException {
          scanner.init(System.in);
          int t = scanner.nextInt();
          while (t-- > 0) {
              int n = scanner.nextInt(), k = scanner.nextInt();
              long x = scanner.nextLong()-1;
              String s = scanner.next();
              List<String> res = new ArrayList<>();
              int i = n-1;
              while(i >= 0) {
                  if(s.charAt(i) == 'a') {
                      res.add("a");
                      i--;
                  }
                  else {
                      int j = i-1;
                      while (j >= 0 && s.charAt(j) == '*') {
                          j--;
                      }
                      int count = (i - j) * k + 1;
                      res.add("b".repeat((int) (x % count)));
                      x /= count;
                      i = j;
                  }
              }
              for(i = res.size()-1; i>=0; i--) {
                  System.out.print(res.get(i));
              }
              System.out.println();
          }
     }
  }
  