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 os.path
  from collections import *
  import math
  import bisect
  import heapq as hq
  from fractions import Fraction
  from random import randint
  
  if os.path.exists("input.txt"):
      sys.stdin = open("input.txt", "r")
      sys.stdout = open("output.txt", "w")
  
  input = sys.stdin.readline
  
  ##########################################################
  def solve():
      n,k,x = [int(x) for x in input().split()]
      s = list(input().strip('\n'))
      y = 0
      h = Counter(s)
      if h['*'] == 0:
          print(''.join(s))
          return
      
      if k == 0:
          for i in range(len(s)):
              if s[i] == '*':
                  s[i] = ''
          print(''.join(s))
          return
  
      s.append('b')
      cnt = 0
      arr = []
      for i in range(len(s)):
          if s[i] == '*':
              cnt += k
          else:
              if cnt:
                  arr.append(cnt)
              cnt = 0
      s.pop()
      prod = 1
      for i in range(len(arr)):
          prod *= (arr[i] + 1)
      
      x = min(x, prod)
      x -= 1
      arr = arr[::-1]
      for i in range(len(arr)):
          arr[i] += 1
          if x == 0:
              arr[i] = 0
              continue
          temp = x % arr[i]
          x //= arr[i]
          arr[i] = temp
  
      arr = arr[::-1]
  
      s1 = []
  
      for i in range(n):
          if s[i] == '*' and (i == 0 or s[i - 1] != '*'):
              s1.append(s[i])
          elif s[i] != '*':
              s1.append(s[i])
      
      k = 0
  
      for i in range(len(s1)):
          if s1[i] == '*':
              s1[i] = arr[k] * 'b'
              k += 1
  
      print(''.join(s1))
  
  t = int(input())
  while t:
      t -= 1
      solve()
  
  
  ##########################################################
  