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: def solver():
      n_1 = int(input())
  
      for _ in range(n_1):
          n, k, x = map(int, input().split())
          f = input()
          a = list(filter(lambda y: y != "", f.split("a")))
  
          list_of_cases = []
          tmp = ""
          for i in f:
              if i == "a":
                  list_of_cases.append(tmp)
                  tmp = ""
                  list_of_cases.append("a")
              else:
                  tmp = tmp + "*"
          list_of_cases.append(tmp)
          list_of_cases = list(filter(lambda y: y != "", list_of_cases))
  
          n_of_cases = 1
          for multiplier in a:
              n_of_cases = n_of_cases * (len(multiplier)*k + 1)
          final = ""
  
          for i in list_of_cases:
              if i == "a":
                  final = final + "a"
              else:
                  length = len(i)
                  chunks = length * k + 1
                  n_of_cases = n_of_cases // chunks
                  for j in range(chunks):
                      if j * n_of_cases < x <= (j+1) * n_of_cases:
                          final = final + j*"b"
                          x = x % n_of_cases
                          if x == 0:
                              x = n_of_cases
  
          print(final)
  
  
  solver()
  