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 {
      public static void main(String[] args) {
          while (N-- > 0) {
              solve();
          }
          out.close();
      }
  
      public static void solve() {
          int n = sc.nextInt();
          int k = sc.nextInt();
          long x = sc.nextLong();
  
          List<Integer> list = new ArrayList<>();
  
          int count_star = 0;
          String s = sc.nextLine();
  
          x--;
  
          for (int i = 0; i < n; i++) {
              if (s.charAt(i) == 'a') {
                  if (count_star != 0) {
                      list.add(count_star * k + 1);
                      count_star = 0;
                  }
                  list.add(0);
              } else {
                  count_star++;
              }
          }
  
          if (count_star != 0) {
              list.add(count_star * k + 1);
          }
  
          for (int i = list.size() - 1; i >= 0 && x > 0; i--) {
              if (list.get(i) == 0) {
                  continue;
              }
  
              int temp = list.get(i);
              list.set(i, -(int) (x % temp) == 0 ? 1 : -(int) (x % temp));
              x /= temp;
          }
  
          for (int i = 0; i < list.size(); i++) {
              if (list.get(i) == 0) {
                  out.print("a");
              } else if (list.get(i) < 0) {
                  for (int j = 0; j < -list.get(i); j++) {
                      out.print("b");
                  }
              }
          }
          out.println();
      }
  
      private static PrintWriter out = new PrintWriter(new BufferedOutputStream(System.out));
  
      private static MyScanner sc = new MyScanner();
  
      private static int N = sc.nextInt();
  
      @SuppressWarnings("unused")
      private static class MyScanner {
          BufferedReader br;
          StringTokenizer st;
  
          public MyScanner() {
              br = new BufferedReader(new InputStreamReader(System.in));
          }
  
          String next() {
              while (st == null || !st.hasMoreElements()) {
                  try {
                      st = new StringTokenizer(br.readLine());
                  } catch (IOException e) {
                      e.printStackTrace();
                  }
              }
              return st.nextToken();
          }
  
          int nextInt() {
              return Integer.parseInt(next());
          }
  
          long nextLong() {
              return Long.parseLong(next());
          }
  
          double nextDouble() {
              return Double.parseDouble(next());
          }
  
          int[] readArray(int n) {
              int[] a = new int[n];
              for (int i = 0; i < n; i++) {
                  a[i] = nextInt();
              }
              return a;
          }
  
          String nextLine() {
              String str = "";
              try {
                  str = br.readLine();
              } catch (IOException e) {
                  e.printStackTrace();
              }
              return str;
          }
      }
  }