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 BAString {
      public static void main(String[] args) throws IOException {
          Reader in = new Reader();
          PrintWriter out = new PrintWriter(System.out);
          
          int T = in.nextInt();
          for (int t = 0; t < T; t++) {
              int N = in.nextInt(), K = in.nextInt();
              long X = in.nextLong() - 1;
              List<Integer> list = new ArrayList<>();
              String s = in.next();
              int len = 0;
              for (int i = 0; i < N; i++) {
                  if (s.charAt(i) == 'a' && i > 0 && s.charAt(i - 1) == '*') {
                      list.add(len);
                      len = 0;
                  }
                  if (s.charAt(i) == '*') {
                      len++;
                  }
              }
              if (len > 0) {
                  list.add(len);
              }
              int M = list.size();
              long[] size = new long[M];
              int[] arr = new int[M];
              for (int i = M - 1; i >= 0; i--) {
                  size[i] = (list.get(i) * K + 1) * (i == M - 1 ? 1 : size[i + 1]);
                  if (size[i] > X || size[i] < 0) {
                      for (int j = i; j < M; j++) {
                          arr[j] = (int)(X / (j == M - 1 ? 1 : size[j + 1]));
                          X %= (j == M - 1 ? 1 : size[j + 1]);
                      }
                      break;
                  }
              }
              StringBuilder res = new StringBuilder("");
              int index = 0;
              for (int i = 0; i < N; i++) {
                  if (s.charAt(i) == '*') {
                      int j = i;
                      while (j + 1 < N && s.charAt(j + 1) == '*') {
                          j++;
                      }
                      for (int k = 0; k < arr[index]; k++) {
                          res.append("b");
                      }
                      i = j;
                      index++;
                  } else {
                      res.append("a");
                  }
              }
              out.println(res);
          }
          
          out.close();
      }
      static class Reader {
          BufferedReader in;
          StringTokenizer st;
          public Reader() {
              in = new BufferedReader(new InputStreamReader(System.in));
              st = new StringTokenizer("");
          }
          public String nextLine() throws IOException {
              st = new StringTokenizer("");
              return in.readLine();
          }
          public String next() throws IOException {
              while (!st.hasMoreTokens()) {
                  st = new StringTokenizer(in.readLine());
              }
              return st.nextToken();
          }
          public int nextInt() throws IOException {
              return Integer.parseInt(next());
          }
          public long nextLong() throws IOException {
              return Long.parseLong(next());
          }
      }
      public static void sort(int[] arr) {
          List<Integer> list = new ArrayList<>();
          for (int i : arr) {
              list.add(i);
          }
          Collections.sort(list);
          for (int i = 0; i < arr.length; i++) {
              arr[i] = list.get(i);
          }
      }
  }