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 defaultdict as dd
  import math
  import sys
  input=sys.stdin.readline
  def nn():
      return int(input())
   
  def li():
      return list(input())
   
  def mi():
      return map(int, input().split())
   
  def lm():
      return list(map(int, input().split()))
   
   
   
  def solve():
   
      n,k,x = mi()
      
      s = input()
      #vim print(s)
      blens =[]
      count =0
      for i in range(len(s)-1):
          
          if s[i]=="*":
              count +=1
          if s[i]=='a' and count>0:
              blens.append(count)
              count = 0
  
      if count>0:
          blens.append(count)
          
      if len(blens)==0:
          print(s[:-1])
          return
      bplace = -1
      count = 0
      bcounts = [0 for i in range(len(blens))]
      bcounts[-1] = (blens[-1])*k+1 
      for i in range(1,len(blens)):
          bcounts[-i-1] = bcounts[-i] * (k * blens[-i-1]+1)
          
      bans = [0 for i in range(len(blens))]
      
      x-=1
      for i in range(1,len(blens)):
          bans[i-1] = x//bcounts[i]
          x = x % bcounts[i]
          
      bans[-1] = x
          
      #print(blens)
      #print(bcounts)
      #print(bans)
      out = []
      count = 0
      place = 0
      for i in range(len(s)-1):
          if s[i] =="a":
              out.append('a')
              count = 0
          if s[i] =="*" and count==0:
              count +=1
              out.append("b"*bans[place])
              place+=1
              
      print("".join(out))
          
              
  q=nn()
  for _ in range(q):
      solve()