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 collections import Counter
  from sys import *
  from collections import defaultdict as dd
  from math import *
  
  # sys.stdin = open('input.txt', 'r')  
  # sys.stdout = open('output.txt', 'w')
  
  #input functions
  def inp():
      return int(stdin.readline().strip())
  def vinp():
      return map(int,stdin.readline().strip().split())
  def linp():
      return list(map(int,stdin.readline().strip().split()))
  def sinp(n = 1):
      if n==1:
          return stdin.readline().strip()
      elif n==2:
          return list(stdin.readline().strip())
      else:
          return list(stdin.readline().split())
  
  #output function
  def pr(*x, end = "\n"):
      print(*x,end = end)
  
  #others
  def mod(f, val = 1000000007):
      return f % val
  def csort(c):
      sorted(c.items(), key=lambda pair: pair[1], reverse=True)
  def indc(l,n):
      c={}
      for i in range(n):
          c[l[i]]=c.get(l[i],[])+[i+1]
      return c
  
  if __name__ =="__main__":
      cou=inp()
      for _ in range(cou):
          n,k,x = vinp()
          x -= 1
          s = sinp()
          s = s[::-1]
          ans = []
          i = 0
          while i < n:
              if s[i] == 'a':
                  ans.append(s[i])
              else:
                  j = i
                  while j + 1 < n and s[j + 1] == s[i]:
                      j += 1
                  temp = (j - i + 1) * k + 1
                  ans.append('b' * (x % temp))
                  x //= temp
                  i = j
              i += 1
          pr(''.join(ans[::-1]))