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: #import io,os
  #input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline
  
  def main(t):
  
  
      n,k,x = map(int,input().split())
      s = input()
  
      digits = []
  
      star = 0
  
  
      for i in range(n):
          if s[i]=="*":  
              star += 1
          else:
              if star>0:  digits.append(star*k+1)
              star = 0
  
      if star>0:  digits.append(star*k+1)
  
  #    print(digits)
      represent = []
  
  
      x -= 1
      while x>0:
          d = digits.pop()
          represent.append(x%d)
          x = x//d
      represent += len(digits)*[0]
    #  represent = represent[::-1]
  
  
  
  #    print(represent)
      ans = []
  
      index = 0
      while index < n:
          if s[index]=='a':
              ans.append('a') 
              index += 1
              continue 
          while index<n and s[index]=='*':
              index += 1
          ans.append(represent.pop()*"b")
  
      ans = "".join(ans)
      
          
          
  
  
  
  
  
  
      print(ans)
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  T = int(input())
  t = 1
  while t<=T:
      main(t)
      t += 1
  