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 os
  import sys
  from io import BytesIO, IOBase
  from types import GeneratorType
  from collections import defaultdict
  BUFSIZE = 8192
  
  def bootstrap(f,stack=[]):
      def fun(*args,**kwargs):
          if stack:
              return f(*args, **kwargs)
          else:
              to = f(*args, **kwargs)
              while True:
                  if type(to) is GeneratorType:
                      stack.append(to)
                      to=next(to)
                  else:
                      stack.pop()
                      if not stack:
                          break
                      to=stack[-1].send(to)
              return to
      return fun
  
  
  class FastIO(IOBase):
      newlines = 0
  
      def __init__(self, file):
          self._fd = file.fileno()
          self.buffer = BytesIO()
          self.writable = "x" in file.mode or "r" not in file.mode
          self.write = self.buffer.write if self.writable else None
  
      def read(self):
          while True:
              b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
              if not b:
                  break
              ptr = self.buffer.tell()
              self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
          self.newlines = 0
          return self.buffer.read()
  
      def readline(self):
          while self.newlines == 0:
              b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
              self.newlines = b.count(b"\n") + (not b)
              ptr = self.buffer.tell()
              self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
          self.newlines -= 1
          return self.buffer.readline()
  
      def flush(self):
          if self.writable:
              os.write(self._fd, self.buffer.getvalue())
              self.buffer.truncate(0), self.buffer.seek(0)
  
  
  class IOWrapper(IOBase):
      def __init__(self, file):
          self.buffer = FastIO(file)
          self.flush = self.buffer.flush
          self.writable = self.buffer.writable
          self.write = lambda s: self.buffer.write(s.encode("ascii"))
          self.read = lambda: self.buffer.read().decode("ascii")
          self.readline = lambda: self.buffer.readline().decode("ascii")
  
  sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
  input = lambda: sys.stdin.readline().rstrip("\r\n")
  import math
  
  # Python program to print all
  # primes smaller than or equal to
  # n using Sieve of Eratosthenes
  
  t=int(input())
  for _ in range(t):
      #n=int(input())
      n,k,x=map(int,input().split())
      #b=list(map(int,input().split()))
      s=input()
      b=[]
      for j in s:
          b.append(j)
  
      res=[]
      curr=0
      for i in range(n):
          if(b[i]=="a"):
              res.append((curr*k)+1)
              curr=0
          else:
              curr+=1
  
      res.append((curr*k)+1)
      ans=[]
      l=len(res)
      j=l-1
      last=1
      req=[]
      while(j>=0):
          if(curr<=x):
              curr=res[j]*last
          else:
              curr=curr
          req.append(curr)
  
          last=curr
          j+=-1
      req.reverse()
  
  
  
      cnt=[0 for i in range(l)]
  
  
      while(x>1):
  
          li=0
          ri=l-1
  
  
          while(li<ri):
              m=(li+ri)//2
  
  
              if(req[m]<x):
                  ri=m-1
              else:
                  if((ri==li+1)):
                      if(req[ri]>=x):
                          li=ri
                      else:
                          ri=li
                  else:
                      li=m
  
          j=li
          if(j==l-1):
              curr=1
          else:
              curr=req[j+1]
          r=x//curr
          if(x%curr==0):
              r-=1
  
          x-=r*curr
  
          cnt[j]=r
  
          last=curr
  
          if(x<=1):
              break
  
  
  
  
      ans=[]
      for i in range(cnt[0]):
          ans.append("b")
  
      for i in range(1,l):
          ans.append("a")
          for j in range(cnt[i]):
              ans.append("b")
  
      print("".join(ans))
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  