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: from __future__ import division, print_function
  from math import inf
  import itertools
  import sys
  
  if sys.version_info[0] < 3:
      input = raw_input
      range = xrange
  
      filter = itertools.ifilter
      map = itertools.imap
      zip = itertools.izip
  
  read_int_vec = lambda x: [int(c) for c in x.split(" ")]
  
  from math import inf, sqrt
  from itertools import groupby
  def solution(s,n,k,x):
      x -= 1
      s = list(s)
      s = s[::-1]
      i = 0
      res = []
      while i < n:
          if s[i] == "a":
              res.append("a")
          else:
              j = i
              while j + 1 < n and s[j+1] == s[i]:
                  j += 1
              curr = (j-i+1) * k + 1
              res.append('b' * (x % curr))
              x //= curr
              i = j
          i += 1
      return "".join(res[::-1])
  
  if __name__ == "__main__":
      """
      test(2,1,7,4, 8, 5, 4, 2)
      test(4,5,7,8, 8, 10, 8, 5)
      test(2,2,5,4, 5, 4, 3, 3)
      test(0,3,1,6, 1, 8, 1, 5)
      """
      n_tests = int(input())
  
      for _ in range(n_tests):
          n,k,x  = read_int_vec(input())
          s = input()
          print(solution(s,n,k,x))