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 BAString {
  	
  	static class FastReader {
  		
          BufferedReader br;
          StringTokenizer st;
   
          public FastReader()
          {
              br = new BufferedReader(
                  new InputStreamReader(System.in));
          }
   
          String next()
          {
              while (st == null || !st.hasMoreElements()) {
                  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()); }
          
          double nextFloat() { return Float.parseFloat(next()); }
   
          double nextDouble() { return Double.parseDouble(next()); }
   
          String nextLine()
          {
              String str = "";
              try {
                  str = br.readLine();
              }
              catch (IOException e) {
                  e.printStackTrace();
              }
              return str;
          }
          
          int[] nextArray(int n)
          {
          	int[] a = new int[n];
          	for (int i=0; i<n; i++) {
          		a[i] = this.nextInt();
          	}
          	return a;
          }
          
          long[] nextArrayLong(int n)
          {
          	long[] a = new long[n];
          	for (int i=0; i<n; i++) {
          		a[i] = this.nextLong();
          	}
          	return a;
          }
          
      }
  	
  	public static void solve(int n, int k, long x, String s) {
  		
  		x--;
  		ArrayList<Long> list = new ArrayList<Long>();
  		for (int i=0; i<n; i++) {
  			long count = 0;
  			while (i <n && s.charAt(i) == '*') {
  				i++;
  				count++;
  			}
  			if (count != 0) {
  				list.add(count * (long) k + 1);
  				i--;
  			}
  		}
  		long[] fin = new long[list.size()];
  		int index = list.size()-1;
  		while (x > 0) {
  			fin[index] = x%list.get(index);
  			x /= list.get(index);
  			index--;
  		}
  		StringBuilder ans = new StringBuilder();
  		index = 0;
  		for (int i=0; i<n; i++) {
  			if (s.charAt(i) == 'a') {
  				ans.append("a");
  			}
  			else {
  				for (int j=0; j<fin[index]; j++) {
  					ans.append("b");
  				}
  				index++;
  				while (i<n && s.charAt(i) == '*') {
  					i++;
  				}
  				i--;
  			}
  		}
  		System.out.println(ans);
  		
  	}
  
  	public static void main(String[] args) {
  		
  		FastReader in = new FastReader();
  		int t = in.nextInt();
  		while (t-- > 0) {
  			solve(in.nextInt(), in.nextInt(), in.nextLong(), in.next());
  		}
  
  	}
  
  }
  