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.*;
  import static java.lang.Math.*;
  
  public class C{
  	static long conv(long i) {
  		if (i == -1) return 1;
  		return i;
  	}
  	public static void main(String[] args) throws IOException {
  		// br = new BufferedReader(new FileReader(".in"));
  		// out = new PrintWriter(new FileWriter(".out"));
  		//new Thread(null, new (), "peepee", 1<<28).start();
  		
  		int t =readInt();
  		while(t-->0) {
  			int n =readInt();
  			long k =readLong();
  			long x =readLong()-1;
  			List<Long> a = new ArrayList<Long>();
  			char[] c = read().toCharArray();
  			long ast = 0;
  			for (int i =0 ; i < n; i++) {
  				if (c[i] == '*') ast++;
  				else {
  					if (ast > 0) a.add(ast*k+1);
  					a.add(-1L);
  					ast = 0;
  				}
  			}
  			if (ast > 0) a.add(ast*k+1);
  			
  			long[] sufprod = new long[a.size()];
  			for (int i = a.size()-1; i >= 0; i--) {
  				sufprod[i] = conv(a.get(i));
  				if (i != a.size()-1) {
  					if (sufprod[i+1] == -1) {
  						sufprod[i] = -1;
  						continue;
  					}
  					if (Long.MAX_VALUE/sufprod[i+1] < sufprod[i]) {
  						sufprod[i] = -1;
  						continue;
  					}
  					sufprod[i]*=sufprod[i+1];
  				}
  			}
  			//if (sufprod[0] < x+1) System.exit(-1);
  			for (int i = 0 ;i < a.size(); i++) {
  				if (a.get(i) == -1) {
  					out.print('a');
  					continue;
  				}
  				if (i == a.size()-1) {
  					while(x-->0) out.print('b');
  					continue;
  				}
  				if (sufprod[i+1] == -1) continue;
  				long suf = sufprod[i+1];
  				long canFit = x/suf;
  				//System.out.println("Canyy " + canFit);
  				x -= canFit*suf;
  				while(canFit-->0) out.print('b');
  				
  			}
  			
  			out.println();
  		}
  		
  		
  		out.close();
  	}
  	
  	/* Stupid things to try if stuck:
  	 * n=1, expand bsearch range
  	 * brute force small patterns
  	 * submit stupid intuition
  	 */
  	
  	static BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
  	static PrintWriter out = new PrintWriter(new BufferedOutputStream(System.out));
  	static StringTokenizer st = new StringTokenizer("");
  	static String read() throws IOException{return st.hasMoreTokens() ? st.nextToken():(st = new StringTokenizer(br.readLine())).nextToken();}
  	static int readInt() throws IOException{return Integer.parseInt(read());}
  	static long readLong() throws IOException{return Long.parseLong(read());}
  	static double readDouble() throws IOException{return Double.parseDouble(read());}
  }