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.*;
  
  public class CF1620C extends PrintWriter {
  	CF1620C() { super(System.out); }
  	Scanner sc = new Scanner(System.in);
  	public static void main(String[] $) {
  		CF1620C o = new CF1620C(); o.main(); o.flush();
  	}
  
  	void main() {
  		int t = sc.nextInt();
  		while (t-- > 0) {
  			int n = sc.nextInt();
  			int k = sc.nextInt();
  			long x = sc.nextLong() - 1;
  			byte[] cc = sc.next().getBytes();
  			int[] aa = new int[n];
  			int[] bb = new int[n];
  			int m = 0;
  			for (int i = 0; i < n; ) {
  				int a = 0;
  				while (i < n && cc[i] == 'a') {
  					a++;
  					i++;
  				}
  				int b = 0;
  				while (i < n && cc[i] != 'a') {
  					b++;
  					i++;
  				}
  				aa[m] = a;
  				bb[m] = b;
  				m++;
  			}
  			for (int h = m - 1; h >= 0; h--) {
  				int b = bb[h] * k + 1;
  				bb[h] = (int) (x % b);
  				x /= b;
  			}
  			StringBuilder sb = new StringBuilder();
  			for (int h = 0; h < m; h++) {
  				while (aa[h]-- > 0)
  					sb.append('a');
  				while (bb[h]-- > 0)
  					sb.append('b');
  			}
  			println(sb);
  		}
  	}
  }
  