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: //package bastring;
  
  import java.util.*;
  import java.io.*;
  
  public class bastring {
  	public static void main(String[] args) throws IOException {
  		BufferedReader fin = new BufferedReader(new InputStreamReader(System.in));
  		int t = Integer.parseInt(fin.readLine());
  		StringBuilder fout = new StringBuilder();
  		while(t-- > 0) {
  			StringTokenizer st = new StringTokenizer(fin.readLine());
  			int n = Integer.parseInt(st.nextToken());
  			long k = Integer.parseInt(st.nextToken());
  			long x = Long.parseLong(st.nextToken()) - 1;
  			String s = fin.readLine();
  			boolean hasA = false;
  			ArrayList<Long> sequence = new ArrayList<Long>();
  			char prev = 'a';
  			for(int i = 0; i < s.length(); i++) {
  				char c = s.charAt(i);
  				if(c == '*') {
  					if(prev == 'a') {
  						sequence.add(k);
  					}
  					else {
  						sequence.set(sequence.size() - 1, sequence.get(sequence.size() - 1) + k);
  					}
  				}
  				prev = c;
  				if(c == 'a') {
  					hasA = true;
  				}
  			}
  			if(!hasA) {
  				x ++;
  			}
  			//System.out.println(sequence);
  			ArrayList<Long> ans = new ArrayList<Long>();
  			for(int i = 0; i < sequence.size(); i++) {
  				ans.add((long) 0);
  			}
  			long mult = 1;
  			int index = 0;
  			outer:
  			for(int i = sequence.size() - 1; i >= 0; i--) {
  				//System.out.println(mult);
  				long count = 1;
  				while(count <= sequence.get(i) + 1) {
  					if(mult * count > x) {
  						index = i;
  						break outer;
  					}
  					count ++;
  				}
  				mult *= sequence.get(i) + 1;
  			}
  			//System.out.println("IN:" + index + " " + x);
  			for(int i = index; i < sequence.size(); i++) {
  				//System.out.println(i + " " + (x / mult));
  				ans.set(i, x / mult);
  				x %= mult;
  				if(i != sequence.size() - 1) {
  					mult /= sequence.get(i + 1) + 1;
  				}
  			}
  			//System.out.println(ans);
  			int counter = 0;
  			outer:
  			for(int i = 0; i < s.length(); i++) {
  				char c = s.charAt(i);
  				if(c == '*') {
  					while(s.charAt(i) == '*') {
  						i ++;
  						if(i == s.length()) {
  							break;
  						}
  					}
  					i--;
  					for(long j = 0; j < ans.get(counter); j++) {
  						fout.append("b");
  					}
  					counter ++;
  				}
  				else {
  					fout.append("a");
  				}
  			}
  			fout.append("\n");
  		}
  		System.out.print(fout);
  	}
  }
  