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 new1{
  
  	
  	
  	
  	public static void main(String[] args)  throws IOException{
      	BufferedWriter output = new BufferedWriter(new OutputStreamWriter(System.out));
  		FastReader s = new FastReader();
  		int t = s.nextInt();
      	for(int z = 0; z < t; z++) {
      		long n = s.nextInt();
      		long k = s.nextInt();
      		long x = s.nextLong() - 1;
      		String str = s.next();
      		ArrayList<Long> aList = new ArrayList<Long>();
      		if(str.charAt(0) == '*') aList.add(k);
      		for(int i = 1; i < n; i++) {
      			if(str.charAt(i) == '*') {
      				if(str.charAt(i - 1) == '*') {
      					long v = aList.get(aList.size() - 1);
      					v = v + k;
      					aList.set(aList.size() - 1, v);
      				}
      				else {
      					aList.add(k);
      				}
      			}
      		}
      		int n1 = aList.size();
      		long[] ac = new long[n1];
      		for(int i = n1 - 1; i >= 0; i--) {
      			long div = aList.get(i) + 1;
      			long val = x % div;
      			x = x / div;
      			ac[i] = val;
      		}
      		StringBuilder ans = new StringBuilder("");
      		int ind = 0;
      		//System.out.println(aList.toString());
      		//System.out.println(Arrays.toString(ac));
      		for(int i = 0; i < n; i++) {
      			if(str.charAt(i) == 'a') {
      				if(i - 1 >= 0  && str.charAt(i - 1) == '*') {
      					for(int j = 0; j < ac[ind]; j++) ans = ans.append("b");
      					ind++;
      				}
      				//System.out.println(i);
      				ans = ans.append("a");
      			}
      		}
      		if(str.charAt((int) (n - 1)) == '*') {
      			for(int j = 0; j < ac[n1 - 1]; j++) ans = ans.append("b");
      		}
      		System.out.println(ans);
      		
      	}
      	//output.flush();
  
      }
  }
  
  
  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();
      }
   
      public int nextInt() {
          return Integer.parseInt(next());
      }
   
      long nextLong() {
          return Long.parseLong(next());
      }
   
      double nextDouble() {
          return Double.parseDouble(next());
      }
   
      String nextLine() {
          String str = "";
          try {
              str = br.readLine();
          } catch (IOException e) {
              e.printStackTrace();
          }
          return str;
  }}