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
  from collections import deque
  import math
  from bisect import bisect_left
  from functools import cmp_to_key
  def II():
  	return int(sys.stdin.readline())
   
  def LI():
  	return list(map(int, sys.stdin.readline().split()))
   
  def MI():
  	return map(int, sys.stdin.readline().split())
   
  def SI():
  	return sys.stdin.readline().strip()
   
  def FACT(n, mod):
      s = 1
      facts = [1]
      for i in range(1,n+1):
          s*=i
          s%=mod
          facts.append(s)
      return facts[n]
   
  def C(n, k, mod):
      return (FACT(n,mod) * pow((FACT(k,mod)*FACT(n-k,mod))%mod,mod-2, mod))%mod
   
  def lcm(a,b):
      return abs(a*b) // math.gcd(a, b)
  
  
  for _ in range(II()):
      n,k,x = MI()
      s = list(SI())
      if s[0] == 'a':
          newS = ['a']
          temp = [0]
      else:
          newS = ['*']
          temp = [k+1]
      for i in range(1,n):
          if s[i] == 'a':
              temp.append(0)
              newS.append('a')
          elif newS[-1] == '*':
              temp[-1]  += k
          else:
              temp.append(k+1)
              newS.append('*')
      d = [0 for i in range(len(s))]
      cur = 0
      start = 0
      s = newS
      for i in range(len(s)-1, -1,-1):
          if cur >= x:
              start = i+1
              break
          if s[i] == '*':
              d[i] = cur
              if cur == 0:
                  cur = temp[i]
              else:
                  cur*=temp[i]
      for i in range(start, len(s)):
          if s[i] == '*':
              if d[i]!=0:
                  tt = math.ceil(x/d[i])-1
                  s[i] = 'b'*tt
                  x -= tt*d[i]
              else:
                  if x == 0:
                      x = temp[i]
                  s[i] = 'b'*(x-1)
                  x = 0
      for i in range(len(s)):
          if s[i] == '*':
              s[i] = ''
      print(''.join(s))