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 sys import stdin, stdout
  rd = lambda: list(map(int, stdin.readline().split()))
  rds = lambda: stdin.readline().rstrip()
  ii = lambda: int(stdin.readline())
  INF = 1 << 62
  mod = 10**9 + 7
  
  
  def solve(s, k, x):
      s = s[::-1] 
      x -= 1
  
      res = []
      bs = []
      i = 0
      while i < len(s):
          if s[i] == 'a':
              res.append('a')
              i += 1
          else:
              curr = 0  # num of '*'
              while i < len(s) and s[i] != 'a':
                  i += 1
                  curr += 1
  
              # index in res, number of bs we can fit here
              bs.append((len(res), curr * k))
              res.append('_')
      
      # decode x
      for j, lim in bs:
          d = x % (lim+1)
          x //= (lim+1)
          res[j] = 'b' * d
  
      return ''.join(res)[::-1]
  
  
  for _ in range(ii()):
      n, k, x = rd()
      s = rds()
  
      print(solve(s,k,x))
  
  #stdout.write(' '.join(map(str, ar)))
  #stdout.write(f'{res}')
  
  