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: t = int(input())
  
  def get_variable_base(n, digit_value):
      current_digit = 0
      suffix_product = get_suffix_product(digit_value)
      based = []
      while current_digit < len(digit_value):
          based.append(n // suffix_product[current_digit + 1])
          n -= suffix_product[current_digit + 1] * based[-1]
          current_digit += 1
      return based
  
  def get_suffix_product(digit_values):
      suffix_product = [1] * (len(digit_values) + 1)
      for i in range(len(suffix_product) - 2, -1, -1):
          suffix_product[i] = suffix_product[i+1] * digit_values[i]
      return suffix_product
  
  def solve():
      n, k, x = map(int, input().split())
      s = input()
      starstrings = list(filter(lambda x: len(x) > 0, s.split('a')))
      astrings = list(filter(lambda x: len(x) > 0, s.split('*')))
      digit_values = list(map(lambda x: len(x) * k + 1, starstrings))
      b_cnt = get_variable_base(x - 1, digit_values)
      b_ptr = 0
      res = ''
      if s[0] == '*':
          res += 'b' * b_cnt[0]
          b_ptr = 1
      for astring in astrings:
          res += astring
          if b_ptr < len(b_cnt):
              res += 'b' * b_cnt[b_ptr]
          b_ptr += 1
      print(res)
  
  for i in range(t):
      solve()