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 sys
  input = sys.stdin.readline
  
  for _ in range(int(input())):
      ans = 0
      n, k, x = map(int,input().split())
      s = input().rstrip()
      
      alist = []
      aster = []
  
      start = s[0]
      if start == "a":
          alist.append(1)
      else:
          aster.append(1)
  
  
      for i in range(1,n):
          if s[i-1] == s[i]:
              if s[i] == "a":
                  alist[-1] +=1
              else:
                  aster[-1] +=1
  
          else:
              if s[i] == "a":
                  alist.append(1)
              else:
                  aster.append(1)
  
  
      if len(aster) == 0:
          print(s)
  
      else:
          for i in range(len(aster)):
              aster[i] *= k
          
          ans_ast = [0 for _ in range(len(aster))]
          ast_val = [0 for _ in range(len(aster))]
  
          ast_val[-1] = 1
          for i in range(len(aster)-2,-1,-1):
              ast_val[i] = ast_val[i+1] * (aster[i+1]+1)
  
          now = x-1
          for i in range(len(aster)):
              ans_ast[i] = now//ast_val[i]
              now = now%ast_val[i]
  
          ans = ""
  
          if start == "a":
              for i in range(len(alist)):
                  ans += "a"*alist[i]
                  try:
                      ans += "b"*ans_ast[i]
                  except:
                      pass
          else:
              for i in range(len(aster)):
                  ans += "b"*ans_ast[i]
                  try:
                      ans += "a"*alist[i]
                  except:
                      pass
          
          print(ans)