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: # Educational Codeforces Round 119 (Rated for Div. 2)
  # C. BA-String
  
  
  def solve():
      n, k, x = list(map(int, input().split()))
      s = input()
  
      if "*" not in s:
          print(s)
          return
  
      stars = [_ for _ in s.split("a") if _]
  
      # A string of q stars can expand to 0, 1, 2, ..., q * k b's, which is
      # 1 + k * q possibilities
      expand = [1 + k * len(star) for star in stars]
  
      # Cumulative product of number of stars
      prods = [expand[-1]]
      for i in range(1, len(stars)):
          prods += [prods[-1] * expand[len(stars) - i - 1]]
  
      x -= 1
  
      # Count how many b's to replace each * by
      counts = []
      for prod in reversed(prods):
          counts += [x // prod]
          x -= prod * (x // prod)
      counts += [x % prods[0]]
  
      globals().update(locals())
  
      # Construct the string
      answer = ""
      i = 1
      j = 0
      while j < len(s):
          if s[j] == "a":
              answer += "a"
              j += 1
          else:
              answer += "b" * counts[i]
              i += 1
              while j < len(s) and s[j] == "*":
                  j += 1
  
      print(answer)
  
      globals().update(locals())
  
  
  for _ in range(int(input())):
      solve()
  