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.*;
  import java.lang.Math;
  public class Solution {
    static long MOD = 1000000007L;
    public static void print(long[] lwler){
    	for(int i=0;i<lwler.length;++i){
    			System.out.print(lwler[i]+" ");
    		}
    		System.out.println();
    }
    public static void print(int[] lwler){
    	for(int i=0;i<lwler.length;++i){
    			System.out.print(lwler[i]+" ");
    		}
    		System.out.println();
    }
    public static void main(String[] args) {
      Scanner in = new Scanner(new BufferedReader(new InputStreamReader(System.in)));
    	int tt = in.nextInt();
    	loop: for(int _m = 0; _m<tt; ++_m){
    		int n = in.nextInt(), k = in.nextInt();
    		long x = in.nextLong();
    		--x;
    		in.nextLine();
    		String S = in.nextLine();
    		String[] toks = S.split("a");
    		//List<Long> l = new ArrayList<>();
    		//for(String ele : toks){
    		//	System.out.println(ele);
    		//}
    		int ct = 0;
    		for(String ele : toks){
    			if(ele.length()>0){
    				++ct;
    				//l.add((ele.length()*(long)k)+1L);
    			}
    		}
    		long[] l = new long[ct];
    		int ii = 0;
    		for(String ele : toks){
    			if(ele.length()>0){
    				l[ii]=(ele.length()*(long)k)+1L;
    				++ii;
    			}
    		}
    		/*for(int i=l.length-2;i>=1;--i){
    			l[i]=l[i]*l[i+1];
    		}*/
    		//print(l);
    		long[] arr = new long[l.length];
    		for(int i=arr.length-1;i>=0;--i){
    			long cur = l[i];
    			arr[i]=(x%cur);
    			x/=l[i];
    		}
    		//print(arr);
    		int ind = 0, p = 0;
    		char[] s = S.toCharArray();
    		StringBuilder sb = new StringBuilder();
    		while(p<s.length){
    			if(s[p]=='a'){
    				sb.append('a');
    				++p;
    				continue;
    			}
    			while(p<s.length&&s[p]!='a'){
    				++p;
    			}
    			for(int i=0;i<arr[ind];++i){
    				sb.append('b');
    			}
    			++ind;
    		}
    		System.out.println(sb.toString());
    	}
    }
  }
  