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.util.*;
  
  public class CF1 {
      public static void main(String[] args) {
          FastScanner sc=new FastScanner();
          int T=sc.nextInt();
  //        int T=1;
          for (int tt=0; tt<T; tt++){
              int n =sc.nextInt();
              int k = sc.nextInt();
              long x = sc.nextLong();
              String s = sc.next();
              int len=0;
              ArrayList<Integer> arr= new ArrayList<>();
              for (int i=0; i<n; i++){
                  if (s.charAt(i)=='a') {if (len!=0) arr.add(len); len=0;}
                  else len++;
              }
              if (len>0) arr.add(len);
  //            if (arr.size()==0|| x==1) System.out.println(s);
  //            else {
                  int section=0;
                  long ans =1;
                  ArrayList<Long> mul= new ArrayList<>();
                  x--;
                  while (section<arr.size()){
                      section++;
                      mul.add(ans);
                      if (x/ans< (long)(arr.get(arr.size()-section)*k+1)) break;
                      ans*= arr.get(arr.size()-section)*k+1;
  //                    System.out.println(ans);
  
                  }
                  int count[]= new int [section];
                  for (int i=0; i<section; i++){
                      count[section-i-1]=  (int)(x/mul.get(section-1-i));
  //                    System.out.println(count[section-i-1]);
                      x%=mul.get(section-1-i);
                  }
                  StringBuilder sb= new StringBuilder();
                  int p= count.length-1;
                  int i= s.length()-1;
                  while (i>=0){
                      if (p>=0 && s.charAt(i)=='*'){
                          while (i>0 && s.charAt(i)=='*' ){
                              i--;
                          }
                          for (int j=0; j<count[count.length-p-1]; j++){
                              sb.append('b');
                          }
                          p--;
  
                      }
                      else if (s.charAt(i)=='a') {sb.append('a');i--;}
                      else i--;
                  }
                  System.out.println(sb.reverse().toString());
  
  //            }
          }
  
      }
  
      static long mod =998244353L;
      static long power (long a, long b){
          long res=1;
          while (b>0){
              if ((b&1)== 1){
                  res= (res * a % mod)%mod;
              }
              a=(a%mod * a%mod)%mod;
              b=b>>1;
          }
          return res;
      }
      boolean[] sieveOfEratosthenes(int n)
      {
  
          boolean prime[] = new boolean[n+1];
          for(int i=0;i<=n;i++)
              prime[i] = true;
  
          for(int p = 2; p*p <=n; p++)
          {
              if(prime[p] == true)
              {
                  for(int i = p*p; i <= n; i += p)
                      prime[i] = false;
              }
          }
          return prime;
      }
      static void sort(int[] a) {
          ArrayList<Integer> l=new ArrayList<>();
          for (int i:a) l.add(i);
          Collections.sort(l);
          for (int i=0; i<a.length; i++) a[i]=l.get(i);
      }
      static void sortLong(long[] a) {
          ArrayList<Long> l=new ArrayList<>();
          for (long i:a) l.add(i);
          Collections.sort(l);
          for (int i=0; i<a.length; i++) a[i]=l.get(i);
      }
      static long gcd (long n, long m){
          if (m==0) return n;
          else return gcd(m, n%m);
      }
  
      static class Pair{
          int x,y;
          public Pair(int x, int y){
              this.x = x;
              this.y = y;
          }
      }
      static class FastScanner {
          BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
          StringTokenizer st=new StringTokenizer("");
          String next() {
              while (!st.hasMoreTokens())
                  try {
                      st=new StringTokenizer(br.readLine());
                  } catch (IOException e) {
                      e.printStackTrace();
                  }
              return st.nextToken();
          }
  
          int nextInt() {
              return Integer.parseInt(next());
          }
          int[] readArray(int n) {
              int[] a=new int[n];
              for (int i=0; i<n; i++) a[i]=nextInt();
              return a;
          }
          long nextLong() {
              return Long.parseLong(next());
          }
      }
  }