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 sys
  import io, os
  input = sys.stdin.readline
  #input = io.BytesIO(os.read(0,os.fstat(0).st_size)).readline
  
  def rl(s):
      if not s:
          return []
      cur = s[0]
      cnt = 0
      res = []
      for c in s:
          if c == cur:
              cnt += 1
          else:
              res.append((cnt, cur))
              cur = c
              cnt = 1
      else:
          res.append((cnt, cur))
      return res
  
  t = int(input())
  for _ in range(t):
      n, k, x= map(int, input().split())
      x -= 1
      s = str(input().rstrip())
  
      nx = [n]*(n+1)
      for i in reversed(range(n)):
          c = s[i]
          if c == 'a':
              nx[i] = i
          else:
              nx[i] = nx[i+1]
      #print(nx)
      #print(s)
      tot = 0
      ans = []
      RL = rl(s)
      M = 1
      #print(RL)
      for i, (cnt, cur) in enumerate(RL):
          if cur == '*':
              M *= cnt*k+1
      #print(M)
      for i, (cnt, cur) in enumerate(RL):
          if cur == 'a':
              ans += ['a']*cnt
          else:
              M //= (cnt*k+1)
              q, x = divmod(x, M)
              ans += ['b']*q
              #print(i, x)
      if x:
          ans += ['b']*x
      #print(x)
      print(''.join(ans))
  