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.PrintStream;
  import java.util.LinkedList;
  import java.util.List;
  import java.util.Scanner;
  
  public class C {
  
      static long MAX = 1000000000000000000L;
  
      public static void main(String[] args) {
          Scanner sc = new Scanner(System.in);
          PrintStream out = System.out;
  
          int t = sc.nextInt();
          for (int i = 0; i < t; i++) {
              int n = sc.nextInt();
              int k = sc.nextInt();
              long x = sc.nextLong();
              String s = sc.next();
              out.println(solve(s, n, k, x));
          }
  
      }
  
      private static String solve(String s, int n, int k, long x) {
          StringBuilder builder = new StringBuilder();
          boolean wasStar = false;
          LinkedList<Integer> kList = new LinkedList<>();
          for (int i = 0; i < s.length(); i++) {
              if (s.charAt(i) == 'a') {
                  builder.append('a');
                  kList.add(0);
                  wasStar = false;
              } else {
                  if (wasStar) {
                      kList.add(kList.removeLast() + k);
                  } else {
                      builder.append('*');
                      kList.add(k);
                  }
                  wasStar = true;
              }
          }
          s = builder.toString();
          int[] ks = kList.stream().mapToInt(v -> v).toArray();
  
          n = ks.length;
          long[] d = new long[n + 1];
          d[n] = 1;
          for (int i = n - 1; i >= 0; i--) {
              if (s.charAt(i) == 'a') {
                  d[i] = d[i + 1];
              } else {
                  if (d[i + 1] == -1) {
                      d[i] = -1;
                      continue;
                  }
                  if ((ks[i] + 1) > MAX / d[i + 1]) {
                      d[i] = -1;
                  } else {
                      d[i] = (ks[i] + 1) * d[i + 1];
                  }
              }
          }
  
          x--;
          StringBuilder res = new StringBuilder();
          for (int i = 0; i < n; i++) {
              if (s.charAt(i) == 'a') {
                  res.append('a');
              } else {
                  if (d[i + 1] != -1) {
                      long bCount = x / d[i + 1];
                      for (int j = 0; j < bCount; j++) {
                          res.append('b');
                      }
                      x -= bCount * d[i + 1];
                  }
              }
          }
  
          return res.toString();
      }
  }
  