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.io.*;
  public class Edu119C{
  	public static void main(String[] main) throws Exception{
  		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
  		StringTokenizer st = new StringTokenizer(br.readLine());
  		PrintWriter out = new PrintWriter(System.out);
  		int t = Integer.parseInt(st.nextToken());
  		for(int trial = 1; trial <= t; trial++){
  			st = new StringTokenizer(br.readLine());
  			int n = Integer.parseInt(st.nextToken());
  			int k = Integer.parseInt(st.nextToken());
  			long x = Long.parseLong(st.nextToken());
  			String s = br.readLine();
  			ArrayList<Integer> consecutiveB = new ArrayList<Integer>();
  			int count = 0;
  			for(int i = 0; i < n; i++) {
  				if(s.charAt(i) == 'a' && count > 0) {
  					consecutiveB.add(count * k);
  					count = 0;
  				}
  				else if(s.charAt(i) == '*')
  					count += 1;
  			}
  			if(count > 0)
  				consecutiveB.add(count * k);
  			long strings = 1;
  			ArrayList<Integer> Bs = new ArrayList<Integer>();
  			for(int i = consecutiveB.size() - 1; i >= 0; i--) {
  				if(consecutiveB.get(i)+1 > (x-1)/strings) {
  					for(int j = 0; j < i; j++)
  						Bs.add(0);
  					int quot = (int)((x-1) / strings);
  					Bs.add(quot);
  					long val = x-1;
  					for(int j = i+1; j < consecutiveB.size(); j++) {
  						val = val % strings;
  						strings /= consecutiveB.get(j) + 1;
  						quot = (int)(val / strings);
  						Bs.add(quot);
  					}
  					break;
  				}
  				strings *= consecutiveB.get(i) + 1;
  			}
  			StringJoiner sj = new StringJoiner("");
  			int index = 0;
  			boolean addB = false;
  			for(int i = 0; i < n; i++) {
  				if(s.charAt(i) == 'a') {
  					if(addB) {
  						for(int j = 0; j < Bs.get(index); j++)
  							sj.add("b");
  						index++;
  						addB = false;
  					}
  					sj.add("a");
  				}
  				else if(s.charAt(i) == '*')
  					addB = true;
  			}
  			if(addB) {
  				for(int j = 0; j < Bs.get(index); j++)
  					sj.add("b");
  			}
  			out.println(sj);
  		}
  		out.close();
  	}
  }