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 itertools import *
  #from math import *
  #from bisect import *
  #from collections import *
  #from random import *
  #from decimal import *
  #from heapq import *
  #from itertools import *            # Things Change ....remember :)
  import sys
  input=sys.stdin.readline
  def inp():
      return int(input())
  def st():
      return input().rstrip('\n')
  def lis():
      return list(map(int,input().split()))
  def ma():
      return map(int,input().split())
  t=inp()
  while(t):
      t-=1
      n,k,x=ma()
      s=st()
      if(x==1):
          for i in s:
              if(i!='*'):
                  print(i,end='')
          print()
          continue
      s=s[::-1]
      sofar=1
      s=list(s)
      ast=0
      s1=[]
      co=0
      for i in range(n):
          if(s[i]=='a'):
              if(ast):
                  s1.append(ast*k + 1)
              s1.append('a')
              ast=0
          else:
              ast+=1
      if(ast):
          s1.append(ast*k + 1)
      #print(s1)
      res=''
      fl=0
      #print(s1)
      for i in range(len(s1)):
          if(fl):
              if(s1[i]=='a'):
                  continue
              else:
                  s1[i]=0
          if(s1[i]=='a'):
              continue
          else:
              if(sofar*s1[i]<x):
                  #res+='b'*s1[i]
                  sofar*=s1[i]
              else:
                  fl=1
                  for j in range(i,-1,-1):
                      if(s1[j]=='a'):
                          continue
                      if(j!=i):
                          sofar=sofar//s1[j]
                      for j1 in range(s1[j]):
                          if((j1+1)*sofar < x):
                              continue
                          s1[j]=j1
                          x-=j1*sofar
                          #print(x,s1,)
                          break
      for i in s1:
          if(i=='a'):
              res+='a'
          else:
              res+='b'*i
      print(res[::-1])
  