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
  
  
  def read_input():
      n, k, x = map(int, stdin.readline().split())
      s = stdin.readline().rstrip()
      return n, k, x, s
  
  
  def solve(n, k, x, s):
      a = [len(si) * k for si in s.split('a') if si != '']
  
      na = [len(si) for si in s.split('*') if si != '']
      nb = []
  
      p = 1
      for i in range(len(a)-1, -1, -1):
          p *= a[i] + 1
          if x <= p:
              x -= 1
              while p > 1:
                  p //= (a[i] + 1)
                  nb.append(x // p)
                  x %= p
                  i += 1
              break
  
      answer = []
      if s[-1] == 'a':
          for nai, nbi in zip(na[::-1], nb[::-1]):
              answer.append(nai * 'a')
              answer.append(nbi * 'b')
          if len(na) > len(nb):
              for nai in na[len(na)-len(nb)-1::-1]:
                  answer.append(nai * 'a')
      else:
          for nai, nbi in zip(na[::-1], nb[::-1]):
              answer.append(nbi * 'b')
              answer.append(nai * 'a')
          if len(na) < len(nb):
              answer.append(nb[0] * 'b')
          elif len(na) > len(nb):
              for nai in na[len(na)-len(nb)-1::-1]:
                  answer.append(nai * 'a')
  
      answer = ''.join(answer[::-1])
      if answer == '':
          answer = 'b'
  
      return answer
  
  
  def main():
      t = int(stdin.readline())
      for _ in range(t):
          input = read_input()
          answer = solve(*input)
          print(answer)
  
  
  if __name__ == '__main__':
      main()
  