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.BufferedReader;
  import java.io.IOException;
  import java.io.InputStreamReader;
  import java.util.ArrayList;
  import java.util.StringTokenizer;
  
  public class CF_1620C {
  	public static void main(String[] args) throws NumberFormatException, IOException {
  		MyReader reader = new MyReader();
  		int t = reader.nextInt();
  		for(int testcase=0; testcase<t; testcase++) {
  			int n = reader.nextInt();
  			int k = reader.nextInt();
  			long x = reader.nextLong()-1;
  			String s = reader.next();
  			ArrayList<Integer> consecAsterisks = new ArrayList<>();
  			int consec = 0;
  			for(int i=0; i<n; i++) {
  				if(s.charAt(i) == '*') consec++;
  				else {
  					consecAsterisks.add(consec);
  					consec = 0;
  				}
  			}
  			consecAsterisks.add(consec);
  			
  			StringBuilder sb = new StringBuilder();
  			long rank = 0;
  			for(int i=0; i<consecAsterisks.size(); i++) {
  				//skip: to avoid overflow, if ways is too large, we won't be able to progress
  				//this is sloppy, so we still need to be careful later. This is just for overflow
  				boolean skip = false;
  				long ways = 1;
  				for(int j=i+1; j<consecAsterisks.size(); j++) {
  					if(2_000_000_000_000_000_000L/ways+2 < consecAsterisks.get(j)*k+1) {
  						skip = true;
  					}
  					else {
  						ways *= consecAsterisks.get(j)*k+1;
  					}
  				}
  				
  				if(skip) {
  					sb.append("a");
  					continue;
  				}
  				
  				
  				long numb = (x-rank)/ways;
  				for(int j=0; j<numb; j++) {
  					sb.append("b");
  				}
  				rank += numb * ways;
  				if(i != consecAsterisks.size()-1) sb.append("a");
  			}
  			System.out.println(sb.toString());
  		}
  	}
  	
  	static class MyReader {
          BufferedReader br;
          StringTokenizer st;
          public MyReader() {
              br = new BufferedReader(new InputStreamReader(System.in));
          }
          public String next() throws IOException  {
              while(st == null || !st.hasMoreElements()) {
                  st = new StringTokenizer(br.readLine());
              }
              return st.nextToken();
          }
          public int nextInt() throws NumberFormatException, IOException {
              return Integer.parseInt(next());
          }
          public long nextLong() throws NumberFormatException, IOException {
              return Long.parseLong(next());
          }
      }
  }
  