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.lang.reflect.Array;
  import java.util.*;
  import java.io.*;
  
  public class A {
      static final long INF = (long) 2e18;
      static ArrayList<Long> al;
  
      public static void main(String[] args) throws IOException {
          Scanner sc = new Scanner(System.in);
          PrintWriter pw = new PrintWriter(System.out);
          int t = sc.nextInt();
          while (t-- > 0) {
              int n = sc.nextInt();
              int k = sc.nextInt();
              long x = sc.nextLong();
              char[] s = sc.next().toCharArray();
              long cnt = 0;
              al = new ArrayList<>();
              for (char c : s) {
                  if (c == 'a') {
                      al.add(cnt);
                      cnt = 0;
                  } else {
                      cnt += k;
                  }
              }
              al.add(cnt);
              long[] cntDP = new long[al.size() + 1];
              cntDP[al.size()] = 1;
              for (int idx = al.size() - 1; idx >= 0; idx--) {
                  try {
                      cntDP[idx] = Math.multiplyExact(cntDP[idx + 1], (al.get(idx) + 1));
                  } catch (Exception e) {
                      cntDP[idx] = INF;
                  }
              }
              StringBuilder sb = new StringBuilder();
              for (int i = 0; i < al.size(); i++) {
                  long h = (x - 1) / cntDP[i + 1];
                  for (int j = 0; j < h; j++) {
                      sb.append('b');
                  }
                  x = x - h * cntDP[i + 1];
                  if (i != al.size() - 1)
                      sb.append('a');
              }
              pw.println(sb);
          }
          pw.close();
      }
  
      static class Scanner {
          BufferedReader br;
          StringTokenizer st;
  
          public Scanner(InputStream s) {
              br = new BufferedReader(new InputStreamReader(s));
          }
  
          public Scanner(FileReader f) {
              br = new BufferedReader(f);
          }
  
          public String next() throws IOException {
              while (st == null || !st.hasMoreTokens())
                  st = new StringTokenizer(br.readLine());
              return st.nextToken();
          }
  
          public int nextInt() throws IOException {
              return Integer.parseInt(next());
          }
  
          public long nextLong() throws IOException {
              return Long.parseLong(next());
          }
  
          public double nextDouble() throws IOException {
              return Double.parseDouble(next());
          }
  
          public int[] nextIntArr(int n) throws IOException {
              int[] arr = new int[n];
              for (int i = 0; i < n; i++) {
                  arr[i] = Integer.parseInt(next());
              }
              return arr;
          }
      }
  }