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 main():
      n, k, x = map(int, input().split())
      s = input()
      
      if len(set(s)) == 1:
          return s 
      
      arr = []
      curr = 0
      for i in range(n):
          if s[i] == '*':
              curr += 1 
          if s[i] == 'a' and curr > 0:
              arr.append(k*curr+1)
              curr = 0
          
      if curr > 0:
          arr.append(k*curr+1)
      # print(arr)
      l = len(arr)
          
      aux = [0]*l
      c = s[0]
      for i in range(1, n):
          if s[i] == '*' and c[-1] != '*':
              c += '*'
          elif s[i] == 'a':
              c += 'a'
        
      # print(c)
      prod = 1 
      
      arr = arr[::-1]
      
      for i in range(l):
          prod = arr[i]*prod 
          # print('yo')
          
          if prod >= x:
              break
          
      extra = arr[:i+1]
      extra = extra[::-1]
      lol = []
      # print(extra)
      for j in range(i+1):
          prod = prod//extra[j]
          
          a = (x-1)//prod
          lol.append(a)
          x -= a*prod
          
      final = [0]*(l-(i+1)) + lol      
      # arr = arr[::-1]
      # blah = []
      # x -= 1 
      # for j in arr:
      #     blah.append(x % j)
      #     x == x // j 
          
      ans = ''
      i = 0
      
      for k in c:
          if k == 'a':
              ans += 'a'
          else:
              for j in range(final[i]):
                  ans += 'b'
              i += 1     
              
          
      
      # print(arr)
      return ans 
      
  for _ in range(int(input())):
      print(main())