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, math
  #sys.setrecursionlimit(10**6)
  INF = float('inf')
  mod = 10**9 + 7
  #mod = 998244353
  input = lambda: sys.stdin.readline().rstrip()
  li = lambda: list(map(int, input().split()))
  
  from itertools import groupby
  
  def RLE(s: str) -> list:
      encoded = []
      for key, group in groupby(s):
          encoded.append([key, len(list(group))])
      return encoded
  
  t = int(input())
  output = []
  for _ in range(t):
      N, K, X = li()
      S = list(input())
      if X == 1:
          cnt = S.count('a')
          output.append('a' * cnt)
          continue
  
      S = RLE(S)
      su = 1
      lst = []
      for k, n in S[::-1]:
          n *= K
          n += 1
          if k == 'a':
              continue
          if n * su >= X:
              su = n * su
              #lst.append(su)
              break
          else:
              su *= n
              lst.append(su)
  
      lst2 = []
      X -= 1
      for i in range(len(lst))[::-1]:
          waru = X // lst[i]
          lst2.append(waru)
          X %= lst[i]
      lst2.append(X)
      
      cnt = len(lst2) - 1
      ans = [''] * len(S)
      for i in range(len(S))[::-1]:
          k, n = S[i]
          if k == 'a':
              ans[i] = 'a' * n
          else:
              if cnt >= 0:
                  ans[i] = 'b' * lst2[cnt]
                  cnt -= 1
  
      output.append(''.join(ans))
  
  
   
  for a in output:
      print(a)