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 C {
  
  	static FastScanner fs = new FastScanner();
  	static Scanner scn = new Scanner(System.in);
  	static PrintWriter pw = new PrintWriter(System.out);
  	static StringBuilder sb = new StringBuilder("");
  	static int INF = 2147483449;
  	
  	public static void main(String[] args) {
  		int t = fs.nextInt();
  		for (int tt = 0; tt < t; tt++) {
  			int n = fs.nextInt(), k = fs.nextInt();
  			long x = fs.nextLong();
  			x--;
  			String s = fs.next();
  			ArrayList<Pair> pairs = new ArrayList<>();
  			pairs.add(new Pair(s.charAt(0), 1));
  			for (int i = 1; i < n; i++) {
  				if (s.charAt(i) == s.charAt(i - 1)) pairs.get(pairs.size() - 1).second++;
  				else pairs.add(new Pair(s.charAt(i), 1));
  			}
  			int m = pairs.size();
  			StringBuilder s2 = new StringBuilder("");
  			for (int i = m-1; i >= 0; i--) {
  				if (pairs.get(i).first == 'a') {
  					for (int j = 0; j < pairs.get(i).second; j++)
  					s2.append('a');
  				}
  				if (pairs.get(i).first == '*') {
  					int p = pairs.get(i).second * k + 1;
  					for (int j = 0; j < x % p; j++)
  						s2.append('b');
  					x /= p;
  				}
  			}
  			for (int i = s2.length() - 1; i >= 0; i--) sb.append(s2.charAt(i));
  			sb.append("\n");
  		}
  		pw.print(sb.toString());
  		pw.close();
  	}
  	
  	static class Pair implements Comparable<Pair> {
  		char first = 0;
  		int second = 0;
  		
  		Pair(char first, int second) {
  			this.first = first;
  			this.second = second;
  		}
  		
  		public int compareTo(Pair o) {
  			if (second == o.second) return Integer.compare(first, o.first);
  			else return Integer.compare(second, o.second);
  		}
  	}
  
  	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) {
  				}
  			return st.nextToken();
  		}
  
  		int nextInt() {
  			return Integer.parseInt(next());
  		}
  
  		long nextLong() {
  			return Long.parseLong(next());
  		}
  
  		double nextDouble() {
  			return Double.parseDouble(next());
  		}
  
  		int[] readArray(int n) {
  			int[] a = new int[n];
  			for (int i = 0; i < n; i++) {
  				a[i] = nextInt();
  			}
  			return a;
  		}
  	}
  }