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 C {
      static FastReader in;
      static PrintWriter out;
  
      public static void main(String[] args) {
          in = new FastReader();
          out = new PrintWriter(System.out);
  
          int t = in.nextInt();
          while (t-- > 0) {
              int n = in.nextInt(), k = in.nextInt();
              long x = in.nextLong();
              x--; // correct division variants 100 / 100 = 1, 99 / 100 = 0
              String s = in.next();
  
              ArrayList<Long> groups = new ArrayList<>();
              int asterisks = 0;
              for (int i = 0; i < s.length(); i++)
                  if (s.charAt(i) == '*')
                      asterisks++;
                  else {
                      if (asterisks != 0)
                          groups.add((long)asterisks * k + 1);
                      asterisks = 0;
                  }
              if (asterisks != 0)
                  groups.add((long)asterisks * k + 1);
              asterisks = 0;
  
              if (groups.size() == 0) {
                  out.println(s);
                  continue;
              }
  
              Long[] variants = new Long[groups.size()];
              for (int i = 0; i < groups.size(); i++)
                  variants[i] = 0L;
              variants[variants.length - 1] = 1L;
              for (int i = variants.length - 2; i >= 0 && Long.MAX_VALUE / groups.get(i + 1) > variants[i + 1]; i-- )
                  variants[i] = groups.get(i + 1) * variants[i + 1];
  
              ArrayList<Long> arrange = new ArrayList<>();
              for (int i = 0; i < variants.length; i++)
                  if (variants[i] == 0) {
                      arrange.add(i, 0L);
                  } else {
                      arrange.add(i, x / variants[i]);
                      x = x % variants[i];
                  }
              asterisks = 0;
              StringBuilder answer = new StringBuilder();
              int arrangeIndex = 0;
              for (int i = 0; i < n; i++)
                  if (s.charAt(i) == '*') {
                      asterisks++;
                      if (asterisks == 1)
                      {
                          for (int j = 0; j < arrange.get(arrangeIndex); j++)
                              answer.append("b");
                          arrangeIndex++;
                      }
                  } else {
                      asterisks = 0;
                      answer.append("a");
                  }
  
              out.println(answer);
          }
  
          out.flush();
          out.close();
      }
  
      static class FastReader {
          BufferedReader br;
          StringTokenizer st;
  
          FastReader() {
              this(System.in);
          }
  
          FastReader(String file) throws FileNotFoundException {
              this(new FileInputStream(file));
          }
  
          FastReader(InputStream is) {
              br = new BufferedReader(new InputStreamReader(is));
          }
  
          int nextInt() {
              return Integer.parseInt(next());
          }
  
          long nextLong() {
              return Long.parseLong(next());
          }
  
          double nextDouble() {
              return Double.parseDouble(next());
          }
  
          String next() {
              while (st == null || !st.hasMoreTokens()) {
                  st = new StringTokenizer(nextLine());
              }
              return st.nextToken();
          }
  
          String nextLine() {
              String line;
              try {
                  line = br.readLine();
              } catch (IOException e) {
                  throw new RuntimeException(e);
              }
              return line;
          }
      }
  }
  