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.*;
  
  public class Main {
  	
  	public static void main(String[] args) throws Exception {
  		BufferedReader in = new BufferedReader(new InputStreamReader(System.in));		
  		BufferedWriter out = new BufferedWriter(new OutputStreamWriter(System.out));		
  		
  		int T = Integer.parseInt(in.readLine());
  		char[] ss = new char[2020];
  		int[] count = new int[2020];
  		
  		long[] dp = new long[2020]; //may overflow
  		
  		for (int ts=1; ts<=T; ts++) {
  			
  			String[] line = in.readLine().split(" ");
  			int N = Integer.parseInt(line[0]);
  			int K = Integer.parseInt(line[1]);
  			long X = Long.parseLong(line[2]);
  			
  			String S = in.readLine();
  			int size = 0;
  
  			{
  				int idx = 0;
  				while (idx < S.length()) {
  					if (S.charAt(idx) == 'a') {
  						ss[size] = 'a';
  						count[size] = 1;
  						size++;
  						idx++;
  						continue;
  					}
  					
  					int left = idx;
  					while (idx < S.length() && S.charAt(idx) == '*') {
  						idx++;
  					}
  					ss[size] = '*';
  					count[size] = idx-left;
  					size++;
  				}
  			}
  			
  			
  			dp[size] = 1;
  			for (int i=size-1; i>=0; i--) {
  				if (ss[i] == 'a') 
  					dp[i] = dp[i+1];
  				else {
  					if (1.0 * dp[i+1] > (1.0 * X / (count[i] * K + 1) + 1e-18)) {
  						dp[i] = X+1;
  					} else {
  						dp[i] = dp[i+1] * (count[i] * K + 1);
  					}
  				}
  			}
  			
  			StringBuilder result = new StringBuilder();
  			for (int i=0; i<size; i++) {
  				if (ss[i] == 'a') {
  					result.append('a');
  					continue;
  				}
  				
  				for (int cnt=0; cnt<=(count[i] * K); cnt++) {
  					if (X > dp[i+1]) {
  						X-=dp[i+1];
  						result.append('b');
  					} else {
  						break;
  					}
  				}
  			}
  			
  			out.write(result.toString() + "\n");
  			out.flush();
  		}
  		
  		in.close();
  		out.close();
  	}
  
  }
  