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.*;
  import java.util.*;
  import java.math.*;
  
  public class Codeforces
  {
      public static void main(String args[])throws Exception
      {
          BufferedReader bu=new BufferedReader(new InputStreamReader(System.in));
          StringBuilder sb=new StringBuilder();
          int t=Integer.parseInt(bu.readLine());
          while(t-->0)
          {
              String s[]=bu.readLine().split(" ");
              int n=Integer.parseInt(s[0]),k=Integer.parseInt(s[1]); long x=Long.parseLong(s[2]);
              char c[]=bu.readLine().toCharArray();
              if(x==0)
              {
                  for(int i=0;i<n;i++)
                  if(c[i]=='a') sb.append('a');
                  sb.append("\n");
                  continue;
              }
  
              int i,l=0;
              ArrayList<Integer> ci=new ArrayList<>();
              for(i=0;i<n;i++)
              if(c[i]=='*') l++;
              else
              {
                  if(l!=0) ci.add(l*k+1);
                  l=0;
              }
              if(l!=0) ci.add(l*k+1);
  
              int sz=ci.size();
              long ri[]=new long[sz+1];
              BigInteger mul=new BigInteger("1");
              for(i=sz-1;i>=0;i--)
              {
                  mul=mul.multiply(mul.valueOf(ci.get(i)));
                  //System.out.println(mul+" "+mul.compareTo(mul.valueOf(x)));
                  if(mul.compareTo(mul.valueOf(x))>0) break;
                  ri[i]=mul.longValue();
              }
              while(i>=0) ri[i--]=(long)2e18;
              ri[sz]=1;
  
              int add[]=new int[sz+1];
              x--;
              while(x>0)
              {
                  int r=rightmost(ri,sz,x);
                  add[r-1]++;
                  x-=ri[r];
              }
  
              StringBuilder sb2=new StringBuilder();
              int j=sz-1;
              for(i=n-1;i>=0;i--)
              {
                  int y=i;
                  while(y>=0 && c[y]=='*') y--;
                  if(y!=i)
                  {
                      while(add[j]-->0) sb2.append('b');
                      i=y+1;
                      j--;
                  }
                  else sb2.append('a');
              }
              sb.append(sb2.reverse()+"\n");
          }
          System.out.print(sb);
      }
  
      static int rightmost(long a[],int r,long x)
      {
          int l=0,mid,ans=r;
          while(l<=r)
          {
              mid=(l+r)>>1;
              if(a[mid]>x)
              {
                  ans=mid+1;
                  l=mid+1;
              }
              else r=mid-1;
          }
          return ans;
      }
  }