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.BufferedReader;
  import java.io.IOException;
  import java.io.InputStreamReader;
  import java.io.PrintWriter;
  import java.math.BigInteger;
  import java.util.*;
  
  public class Main {
  
      BufferedReader in;
      PrintWriter out;
  
      Main() {
          in = new BufferedReader(new InputStreamReader(System.in));
          out = new PrintWriter(System.out);
      }
  
      void close() throws IOException {
          in.close();
          out.flush();
          out.close();
      }
  
      void solve1() throws IOException {
          int t = Integer.parseInt(in.readLine());
          while (t > 0) {
              StringTokenizer st = new StringTokenizer(in.readLine());
              int n = Integer.parseInt(st.nextToken());
              int k = Integer.parseInt(st.nextToken());
              long x = Long.parseLong(st.nextToken());
              String s = in.readLine();
              List<Integer> bs = new ArrayList<>();
              StringBuilder sb = new StringBuilder();
              int p = 0;
              for (int i = 0; i < n; i++) {
                  char c = s.charAt(i);
                  if (c == 'a') {
                      if (p > 0) {
                          sb.append('*');
                          bs.add(p * k);
                          p = 0;
                      }
                      sb.append('a');
                  } else if (c == '*') {
                      p++;
                  }
              }
              if (p > 0) {
                  sb.append('*');
                  bs.add(p * k);
              }
              String ns = sb.toString();
              int m = bs.size();
              BigInteger[] ps = new BigInteger[m];
              BigInteger q = BigInteger.ONE;
              for (int i = m - 1; i >= 0; i--) {
                  ps[i] = q;
                  q = q.multiply(BigInteger.valueOf(bs.get(i) + 1));
              }
              int[] as = new int[m];
              x--;
              for (int i = 0; i < m; i++) {
                  BigInteger bx = BigInteger.valueOf(x);
                  if (bx.compareTo(ps[i]) < 0) {
                      continue;
                  }
                  as[i] = (int) (x / ps[i].longValue());
                  x %= ps[i].longValue();
                  if (x == 0) {
                      break;
                  }
              }
              p = 0;
              StringBuilder nsb = new StringBuilder();
              for (int i = 0; i < ns.length(); i++) {
                  char c = ns.charAt(i);
                  if (c == '*') {
                      for (int j = 0; j < as[p]; j++) {
                          nsb.append('b');
                      }
                      p++;
                  } else {
                      nsb.append(c);
                  }
              }
              out.append(String.format("%s\n", nsb));
              t--;
          }
      }
  
      String toStr(int n, String s, int[] bs) {
          StringBuilder sb = new StringBuilder();
          for (int i = 0; i < n; i++) {
              char c = s.charAt(i);
              if (c == 'a') {
                  sb.append(c);
              } else if (c == '*') {
                  if (bs[i] > 0) {
                      for (int j = 0; j < bs[i]; j++) {
                          sb.append('b');
                      }
                  }
              }
          }
          return sb.toString();
      }
  
      void permutation(String s, int[] bs, int i, int k, int n) {
          if (i == n) {
              String ns = toStr(n, s, bs);
              set.add(ns);
              return;
          }
          char c = s.charAt(i);
          if (c == 'a') {
              permutation(s, bs, i + 1, k, n);
          } else if (c == '*') {
              for (int j = 0; j <= k; j++) {
                  bs[i] = j;
                  permutation(s, bs, i + 1, k, n);
              }
          }
      }
  
      Set<String> set;
  
      void solve() throws IOException {
          int t = Integer.parseInt(in.readLine());
          while (t > 0) {
              StringTokenizer st = new StringTokenizer(in.readLine());
              int n = Integer.parseInt(st.nextToken());
              int k = Integer.parseInt(st.nextToken());
              long x = Long.parseLong(st.nextToken());
              String s = in.readLine();
              set = new HashSet<>();
              int[] bs = new int[n];
              permutation(s, bs, 0, k, n);
              List<String> list = new ArrayList<>();
              list.addAll(set);
              Collections.sort(list);
              out.append(String.format("%s\n", list.get((int) x - 1)));
              t--;
          }
      }
  
      public static void main(String[] args) throws IOException {
          Main m = new Main();
          m.solve1();
          m.close();
      }
  
  }