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 merge(string, k):
      arr = []
      mrgstring = []
      prevstar = False
      temp = 0
      for i in range(1, len(string)+1):
          
          if string[-1*i] == '*':
              prevstar = True
              temp +=1
          else:
              if prevstar:
                  mrgstring.append('*')
                  arr.append(k * temp + 1)
                  temp = 0
                  prevstar = False
              mrgstring.append('a')
      if prevstar:
          mrgstring.append('*')
          arr.append(k * temp + 1)
      #print('arr= ', arr)
      mrgstring.reverse()
      mrgstring = ''.join(mrgstring)
      #arr.reverse()
      return arr, mrgstring
  
  def f(string ,arr, x):
      farr = []
      x= x-1
      #print(string, arr, x)
      for i in arr:
          if i == 0:
              farr.append(0)
          else:
              farr.append(x%i)
              x= int(x//i)
      #print(string, farr, x)
      farr.reverse()
      temp= ''
      j = 0
      for i in string:
          if i == '*':
              temp+= 'b'* int(farr[j])
              j+=1
          else:
              temp+='a'
      print(temp)
  
  t = int(input())
  for i in range(t):
      n , k, x = list(map(int, input().split(' ')))
      q = input()
      arr , mstr = merge(q, k)
      f(mstr, arr, x)
  
  
  