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.*;
  
  public class grT {
      static String star(ArrayList<String> str,ArrayList<String> ar, int k, long x){
          int l = ar.size()-1;
          int tmp=0;
          Stack<Long> d =new Stack<>();
          Stack<Long> ml = new Stack<>();
          String s="";
          //System.out.println("1");
          int o = ar.get(l).indexOf('*');
              if((ar.get(l).length()-o)*k+1>=x){
                  tmp = Integer.valueOf(ar.get(l).substring(0,o));
                  str.set(tmp,"b".repeat((int)x-1));
                   }else{
                  long tot = (ar.get(l).length()-o)*k+1;
                  ml.add((long) tot);
                  while(true){
                      l--;
                      o = ar.get(l).indexOf('*');
                      if(tot*((ar.get(l).length()-o)*k+1)>=x||tot*((ar.get(l).length()-o)*k+1)<0){
                          l++;
                          o = ar.get(l).indexOf('*');
                          tot = tot*((ar.get(l).length()-o)*k+1);
                          break;
                      }else{
                          tot = (tot)*((ar.get(l).length()-o)*k+1);
                          ml.add((long) tot);
                      }
                  }
                  //System.out.println(ml);
                  while(l<ar.size()){
                      o = ar.get(l).indexOf('*');
                      //tot = tot/((ar.get(l).length()-o)*k+1);
                      if(x%ml.peek()==0){
                          d.add(x/ml.peek()-1);
                          x=ml.peek();
                      }
                      else {
                          d.add((x/ml.peek()));
                          x = x % ml.peek();
                      }
                      ml.pop();
                      l++;
                      if(l==ar.size()){
                          d.add((x-1));
                      }
                      //System.out.println(d);
                  }
                  //System.out.println(d);
                  l=ar.size()-1;
                  while(l>=0){
                      o = ar.get(l).indexOf('*');
                      tmp = Integer.valueOf(ar.get(l).substring(0,o));
                      //System.out.println(d.peek());
                      str.set(tmp,"b".repeat(Math.toIntExact(d.pop())));
                      //System.out.println(4);
                      l--;
                      if(d.size()==0)
                          break;
                  }
              }
              for(int i =0; i<str.size(); i++) {
                  if(str.get(i).indexOf('*')<0)
                      s += str.get(i);
              }
              return s;
      }
      public static void main(String[] args) {
          Scanner sc= new Scanner(System.in);
          int t  = sc.nextInt();
          while(t-->0){
              int n = sc.nextInt();
              int k = sc.nextInt();
              long x = sc.nextLong();
              String s=sc.next();
              ArrayList<String> ar = new ArrayList<>();
              String a="0";
              ArrayList<String> str = new ArrayList<>();
              int ind =0;
              for(int i=0; i<n; i++){
                  if(s.charAt(i)=='*'){
                      a+="*";
                  }else{
                      if(a.indexOf("*")>=0){
                          ar.add(a);
                          str.add(a);
                          ind++;
                      }
                      a=String.valueOf(ind+1);
                      str.add(String.valueOf(s.charAt(i)));
                      ind++;
                  }
              }
              if(a.indexOf("*")>=0){
                  ar.add(a);
                  str.add(a);
              }
              if(ar.size()==0){
                  System.out.println(s);
                  continue;
              }
              if(x==1||k==0){
                  s="";
                  for(int i =0; i<str.size(); i++) {
                      if(str.get(i).indexOf('*')<0){
                          s += str.get(i);
                      }
                  }
                  System.out.println(s);
                  continue;
              }
              s = star(str,ar,k,x);
              System.out.println(s);
          }
      }
  }
  