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: def fun(arr, x):
      y = 1
      i = 0
      while y < x:
          y *= arr[i]
          i += 1
      i -= 1
      y = y//arr[i]
      z = y
      y = 0
      j = 0
      while y < x:
          y += z
          j += 1
      j-= 1
      return [i,j,y-z]
  
  for _ in range(int(input())):
      n, k, x = map(int,input().split())
      s = input()
      arr = []
      c = 0
      flag = True
      for i in s:
          if i == 'a':
              if flag:
                  arr.append(c*k+1)
                  c = 0
                  flag = False
          else:
              c += 1
              flag = True
      if flag:
          arr.append(c*k+1)
      if s[0] == 'a':
          arr.pop(0)
      arr = arr[::-1]
      ans = [0]*len(arr)
      sol = ''
      while x > 1:
          coord = fun(arr, x)
          i, j = coord[0], coord[1]
          # print(coord)
          x -= coord[2]
          # print(x)
          ans[i] = j
      ans = ans[::-1]
  
      flag = True
      j = 0
      for i in s:
          if i == 'a':
              sol += 'a'
              flag = True
          else:
              if flag:
                  sol += 'b'*ans[j]
                  j += 1
                  flag = False
      if sol == '':
          print('b')
      else:
          print(sol)