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 C {
      static MyScanner sc;
      static PrintWriter out;
      static {
          sc = new MyScanner();
          out = new PrintWriter(System.out);
      }
      public static ArrayList<Pair> run_encode(char[] a) {
          int n = a.length;
          ArrayList<Pair> ans = new ArrayList<>();
          char prev = a[0];
          int c = 1;
          for(int i = 1; i < n; i++) {
              if(a[i] == prev) c++;
              else{
                  ans.add(new Pair(prev, c));
                  c = 1;
                  prev = a[i];
              }
          }
          ans.add(new Pair(prev, c));
          return ans;
      }
      public static void solve() {
          int n = sc.nextInt();
          int k = sc.nextInt();
          long x = sc.nextLong() - 1;
          char[] a = sc.next().toCharArray();
          StringBuilder ans = new StringBuilder();
          ArrayList<Pair> A = run_encode(a);
          for(int i = A.size() - 1; i >= 0; i--) {
              Pair p = A.get(i);
              if(p.x == 'a')
                  p.res = p.y;
              else {
                  p.y = p.y * k + 1;
                  p.res = x % p.y;
                  x = x / p.y;
              }
          }
          for(Pair p : A) {
              for(int i = 0; i < p.res; i++)
                  out.print((p.x == 'a' ? 'a' : 'b'));
          }
          out.println();
      }
      public static void main(String[] args) {
          int t = sc.nextInt();
          while(t-- > 0)
              solve();
          out.flush();
      }
  }
  
  class Pair {
      char x;
      int y;
      long res;
      Pair(char a, int b) {
          x = a;
          y = b;
          res = 0;
      }
      public String toString() {
          return "(" + x +", " + y + ", " + res + ")";
      }
  }
  
  class MyScanner {
      BufferedReader br;
      StringTokenizer tok;
      MyScanner() {
          try { br = new BufferedReader(new InputStreamReader(System.in)); }
          catch(Exception e) { System.out.println(e); }
          tok = new StringTokenizer("");
      }
      public String next() {
          try {
              while(!tok.hasMoreTokens()) tok = new StringTokenizer(br.readLine());
          }
          catch(Exception e) { System.out.println(e); }
          return tok.nextToken();
      }
      public int nextInt() {
          return Integer.parseInt(next());
      }
      public long nextLong() {
          return Long.parseLong(next());
      }
      public double nextDouble() {
          return Double.parseDouble(next());
      }
  }
  