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
  input = sys.stdin.readline
  def iinput(): return int(input())
  def sinput(): return input().rstrip()
  def i0input(): return int(input()) - 1
  def linput(): return list(input().split())
  def liinput(): return list(map(int, input().split()))
  def miinput(): return map(int, input().split())
  def li0input(): return list(map(lambda x: int(x) - 1, input().split()))
  def mi0input(): return map(lambda x: int(x) - 1, input().split())
  INF = 10**20
  MOD = 1000000007
  
  def rle(s):
      tmp, count, ans1, ans2 = s[0], 1, [], []
      for i in range(1,len(s)):
          if tmp == s[i]:
              count += 1
          else:
              ans1.append(tmp)
              ans2.append(count)
              tmp = s[i]
              count = 1
      ans1.append(tmp)
      ans2.append(count)
      return ans1, ans2
  
  t = iinput()
  for _ in [0] * t:
      n, k, x = liinput()
      x -= 1
      s = sinput()
      tmp = s.count('*')
      if tmp == 0:
          print(s)
          continue
      ss, ctr = rle(s)
      res = []
      for c, cn in zip(ss[::-1], ctr[::-1]):
          if c == 'a':
              continue
          else:
              res.append(x % (k * cn + 1))
              x //= k * cn + 1
      ans = ''
      for c, cn in zip(ss, ctr):
          if c == 'a':
              ans += 'a' * cn
          else:
              tmp = res.pop()
              ans += 'b' * tmp
      print(ans)