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.io.PrintWriter;
  import java.util.ArrayList;
  import java.util.StringTokenizer;
   
  public class taskc {
  	public static void main(String[] args) {
  		FastScanner in = new FastScanner();
  		PrintWriter out = new PrintWriter(System.out);
  		int tt = in.nextInt();
  		while (tt-- > 0) {
  			int n = in.nextInt(), k = in.nextInt();
  			long x = in.nextLong();
  			x--;
  			String s = in.next();
  			ArrayList<Seg> a = new ArrayList<>();
  			for (int i = 0; i < n; i++) {
  				if (a.isEmpty() || a.get(a.size() - 1).c != s.charAt(i)) {
  					a.add(new Seg(s.charAt(i), 1));
  				} else {
  					a.get(a.size() - 1).cnt++;
  				}
  			}
  			StringBuilder ans = new StringBuilder("");
  			for (int i = a.size()-1; i >= 0; i--) {
  				if (a.get(i).c == '*') {
  					int len = a.get(i).cnt * k;
  					for (int j = 0; j < x % (len + 1); j++) {
  						ans.append('b');
  					}
  					x /= (len + 1);
  				} else {
  					for (int j = 0; j < a.get(i).cnt; j++) {
  						ans.append('a');
  					}
  				}
  			}
  			ans.reverse();
  			out.println(ans);
  		}
  		out.flush();
  	}
  
  	static class Seg {
  		char c;
  		int cnt;
  		Seg(char c, int cnt) {
  			this.c = c;
  			this.cnt = cnt;
  		}
  	}
  
  	static class FastScanner {
  		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
  		StringTokenizer st = new StringTokenizer("");
  		String next() {
  			while (!st.hasMoreTokens()) {
  				try {
  					st = new StringTokenizer(br.readLine());
  				} catch (IOException e) {
  					e.printStackTrace();
  				}
  			}
  			return st.nextToken();
  		}
  		int nextInt() { return Integer.parseInt(next()); }
  		long nextLong() { return Long.parseLong(next()); }
  		float nextFloat() { return Float.parseFloat(next()); }
  		double nextDouble() { return Double.parseDouble(next()); }
  	}
  }