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
   
  BUFSIZE = 8192
   
   
  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")
  #######################################
  
  a=int(input())
  for i in range(a):
      n,k,x=map(int,input().split())
      s=input()
      
      x-=1
      base=k
      cnt=0
      flag=1
  
      ans=[]
      total=0
      ori=[]
      g1=1
  
      gt=0
      base=k
      q=pow(10,20)
      for i in range(len(s)-1,-1,-1):
          gt=min(gt,q)
          base=min(base,q)
          g1=min(g1,q)
          if(s[i]=='*' and flag==0):
              ans.append([i,min(base,q)])
              ori.append(g1)
              gt+=base
              
          if(s[i]=='a' and flag==0):
              
              
              base=(gt+1)*k
              
              g1=gt+1
              flag=1
             
              
              
              
  
              
              
              
          if(s[i]=='a' and flag==1):
              continue
          if(s[i]=='*' and flag==1):
              gt+=base
              ans.append([i,base])
              ori.append(g1)
              flag=0
      ans=ans[::-1]
      ori=ori[::-1]
      
      g1=[]
      m3=[]
      for i in range(len(ans)):
          idx=ans[i][0]
          
          if(x>=ans[i][1]):
              g1.append([idx,k])
              x-=ans[i][1]
              continue
          elif(x<ori[i]):
              g1.append([idx,0])
              continue
          m=x//ori[i]
          g1.append([idx,m])
          x-=m*ori[i]
  
      fin=[]
      i1=0
      for i in range(len(s)):
          if(s[i]=='a'):
              fin.append('a')
          else:
          
              for j in range(g1[i1][1]):
                  fin.append('b')
              i1+=1
    
      print(''.join(map(str,fin)))
              
              
              
  
          
              
              
  
  
     
  