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.math.BigInteger;
  import java.util.*;
  
  public class Div21620C {
  
  	public static void main(String[] args) throws IOException {
  
  		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
  		PrintWriter pw = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
  		
  		int t = Integer.parseInt(br.readLine());
  		
  		while(t --> 0) {
  						
  			StringTokenizer st = new StringTokenizer(br.readLine());
  			
  			int n = Integer.parseInt(st.nextToken());
  			BigInteger k = BigInteger.valueOf(Integer.parseInt(st.nextToken()));
  			BigInteger x = BigInteger.valueOf(Long.parseLong(st.nextToken()));
  						
  			char[] line = br.readLine().toCharArray();
  			
  			BigInteger mult = BigInteger.ONE;
  			BigInteger cur = BigInteger.ZERO;
  						
  			ArrayList<BigInteger> list = new ArrayList<BigInteger>();
  			ArrayList<BigInteger> listC = new ArrayList<BigInteger>();
  			
  			for(int i = n-1; i >= 0; i--) {
  								
  				if(line[i] == '*') {
  										
  					cur = cur.add(k);
  																													
  				}else if(cur.compareTo(BigInteger.ZERO) > 0) {
  					
  					list.add(cur);
  					listC.add(mult);
  					
  					mult = mult.multiply(cur.add(BigInteger.ONE));				
  					cur = BigInteger.ZERO;
  															
  				}
  																				
  			}	
  			
  			if(cur.compareTo(BigInteger.ZERO) > 0) {
  				
  				list.add(cur);
  				listC.add(mult);
  				
  			}
  			
  			StringBuilder sb = new StringBuilder("");
  			int ind = list.size()-1;
  			
  			for(int i = 0; i < n; i++) {
  												
  				if(line[i] == 'a') {
  										
  					sb.append('a');
  					
  				}else if(ind >= 0){
  													
  					for(BigInteger j = list.get(ind); j.compareTo(BigInteger.ZERO) >= 0; j = j.subtract(BigInteger.ONE)) {
  												
  						if(j.multiply(listC.get(ind)).compareTo(x) < 0) {
  															
  							for(int g = 0; g < j.longValue(); g++)
  								sb.append('b');
  							
  							x = x.subtract(j.multiply(listC.get(ind)));
  							i += list.get(ind).divide(k).longValue()-1;
  							ind--;
  							
  							break;
  																					
  						}
  						
  					}
  									
  				}
  				
  			}
  			
  			pw.println(sb);
  			
  		}
  		
  		pw.close();
  		
  	}
  
  }
  