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.*;
  import java.util.Scanner;
  
  
  public class Solution {
  	static int mod=1000000007;;
  	
  //	 
  	public static void main(String[] args) {
  		// TODO Auto-generated method stub
  //		System.out.println();
  
  		Scanner sc=new Scanner(System.in);
  		int tt=sc.nextInt();
  		
  		
  		while(tt-->0){
  			
  //
  //		
  			int n=sc.nextInt();
  			long k=sc.nextLong();
  			long x=sc.nextLong();
  			String s=sc.next();
  			
  			long steps=1;
  			int j=n-1;
  			
  			long arr[]=new long[n];
  			arr[n-1]=1;
  			while(j>=0&&s.charAt(j)!='*') {
  				arr[j]=1;
  				j--;}
  			
  			boolean no=true;
  			for(int i=0;i<n;i++) {
  				if(s.charAt(i)=='a') {no=false;
  				break;}
  			}
  			
  			if(no)x++;
  			
  			while(j>=0&&steps<x) {
  				
  				
  				if(s.charAt(j)=='*') {
  					int cont=0;
  					while(j>=0&&s.charAt(j)=='*') {
  						arr[j]=steps;
  						j--;
  					
  					cont++;}
  					
  					long curr=cont*k;
  					steps+=curr*steps;
  					if(steps<0)break;
  					
  					
  					
  				}
  				else {
  					
  					j--;
  				}
  				
  				
  				
  				
  			}
  			
  			
  			
  			
  			
  			
  			
  			j++;
  			StringBuilder st=new StringBuilder();
  			for(int i=0;i<j;i++) {
  				if(s.charAt(i)=='a') {st.append('a');}
  			}
  			for(int i=j;i<n;i++) {
  			
  				if(s.charAt(i)=='a') {st.append('a');
  				continue;}
  							
  
  				long c=(x-1)/arr[i];
  //				System.out.println(c+" "+st+" "+x);
  				
  				for(long l=0;l<c;l++)st.append('b');
  				
  				x-=c*arr[i];
  				
  				while(i<n&&s.charAt(i)=='*') {i++;
  					}
  				i--;
  				
  			}
  //			
  //			for(long i:arr)System.out.print(i+" ");
  			
  			System.out.println(st.toString());
  			
  		}
  			
  			
  			
  				
  				
  			
  		
  			
  		
  			
  		}
  		
  		
  		
  		
  		
  		
  	
  	
  	
  
  }