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.*;
  public class Solution {
      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 nextDouble() {
              return Double.parseDouble(next());
          }
          String nextLine() {
              String str = "";
              try {
                  str = br.readLine();
              } catch (IOException e) {
                  e.printStackTrace();
              }
              return str;
          }
      }
      static void sort(int a[]){ // int -> long
          ArrayList<Integer> arr=new ArrayList<>(); // Integer -> Long
          for(int i=0;i<a.length;i++)
          arr.add(a[i]);
          Collections.sort(arr);
          for(int i=0;i<a.length;i++)
          a[i]=arr.get(i);
          
      }
      private static long gcd(long a, long b){
          if(b==0)return a;
          return gcd(b,a%b);
      }
      private static long pow(long x,long y){
          if(y==0)return 1;
          long temp = pow(x, y/2);
          if(y%2==1){
              return x*temp*temp;
          }
          else{
              return temp*temp;
          }
      }
      static int log(long n){
          int res = 0;
          while(n>0){
              res++;
              n/=2;
          }
          return res;
      }
  
      static int mod = (int)1e9+7;
      static PrintWriter out;
      static FastReader sc ;
      public static void main(String[] args) throws IOException {
          sc = new FastReader();
          out = new PrintWriter(System.out);
          // primes();
          // ================================ //
          int test = sc.nextInt();
          while (test-- > 0) {
              int n= sc.nextInt();
              int k= sc.nextInt();
              long x= sc.nextLong();
              String s = sc.nextLine();
              solver(s, n, k, x);
          }
          // ================================ //
          // int n = sc.nextInt();
          // solver();
          // ================================ //
          out.flush();
      }
  
      public static void solver(String s, int n, int k, long x) {
          x--;
          int[] count = new int[n];
          int pos = 0;
          for(int i=0;i<n;i++){
              if(s.charAt(i)=='*')count[pos]++; 
              else if(count[pos]>0) pos++;
          }
          
          int len=count[pos]==0?pos-1:pos;
          pos = len;
          int[] map = new int[n];
          while(x>0){
              // System.out.println("__"+ x);
              map[pos] = (int)(x%(1+k*count[pos]));
              x/=(count[pos]*k+1);
              pos--;
          }
          StringBuilder str = new StringBuilder("");
          pos = 0;
          for(int i=0;i<n;i++){
              if(s.charAt(i)=='*'){
                  str.append("b".repeat(map[pos])); 
                  while(i<n-1 && s.charAt(i+1)=='*'){
                      i++;
                  }
                  pos++;
              }
              else{
                  str.append(s.charAt(i));
              }
          }
          out.println(str);
      }
  }
  