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
  input = stdin.readline
  
  from math import ceil
  
  def get(x):
  
      x_dup = x
      value = 1
      for i in range(len(take) - 1 , -1 , -1):
  
          if(x <= (value * (take[i] - (i != len(take) - 1)))):
              
              this = ceil(x/value)
              vals[i] = this - (i == len(take) - 1)
              x_dup = x_dup - (this * value)
              break
  
          x -= (value * (take[i] - (i != len(take) - 1)))
          value *= take[i]
  
      return x_dup
  
  def answer(x):
      global vals
  
      if(len(take) == 0):return a
      
      while(x):
          x = get(x)
  
      ans , i , ind = "" , 0 , 0
      while(i < n):
  
          if(a[i] == 'a'):
              ans += 'a'
              i += 1
          else:
              ans += 'b'*vals[ind]
              while(i < n and a[i] == '*'):i += 1
              ind += 1
  
      return ans
  
  for T in range(int(input())):
  
      n , k , x = map(int,input().split())
      a = input().strip()
  
      take , count = [] , 0
      for i in range(n):
          if(a[i] == '*'):count += 1
          else:
              if(count):take.append((count * k) + 1)
              count = 0
  
      if(count):take.append((count * k) + 1)
  
      vals = [0] * len(take)
      
      print(answer(x))
  