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 = lambda: sys.stdin.readline().rstrip()
  INF = 10 ** 19
  
  def solve():
      n, k, x = map(int,input().split())
  
      s = list(input())
  
      ren = []
      cnt = 0
      for i in range(n):
          if s[i] == '*':
              cnt += 1
          else:
              ren.append(cnt)
              cnt = 0
      t = []
      for i in range(n):
          if s[i] == 'a':
              t.append(s[i])
          elif s[i] == '*' and i == n - 1:
              t.append('*')
          elif s[i] == '*' and s[i + 1] == 'a':
              t.append('*')
      s = t
      n = len(s)
      ren.append(cnt)
      l = []
      for i in ren:
          if i > 0:
              l.append(i * k)
      ans = []
      x -= 1
      for i in range(len(l) - 1, -1, -1):
          ans.append(x % (l[i] + 1))
          x //= (l[i] + 1)
  
      for i in range(n):
          if s[i] == '*':
              if not ans:
                  s[i] = ''
                  continue
              s[i] = 'b' * (ans.pop())
      print(''.join(s))
  
      
  
  
  
  
              
  
  t = int(input())
  for _ in range(t):
      solve()