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: t=int(input())
  for _ in range(t):
      # try:
      n,k,x = map(int,input().split())
      np,kp,xp=n,k,x
      s = list(input())
      arr = []
      i=0
      sp = s[:]
      if(x==1):
          s="".join(s)
          s=s.replace("*","")
          print(s)
          continue
      x-=1
      while(i<len(s)):
          if(s[i]!='*'):
              arr+=[s[i]]
              i+=1
          else:
              count=1
              while(i<len(s) and s[i]=='*'):
                  i+=1
                  count+=1
              arr+=[(count-1)*k]
      i = len(arr)-1 
      remaining = x
      while(i>=0):
          if(arr[i]=='a'):
              i-=1
              continue
          nn = arr[i]
          kk = remaining % (nn+1)
          arr[i] = 'b'*int(kk)
          remaining=remaining//(nn+1)
          if(remaining==0):
              break
          i-=1
      arr = [item for item in arr if not isinstance(item,int)]
      print("".join(arr))
      # except:
      #     print(np,kp,xp,sp)