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: '''
      Python3(PyPy3) Template for Programming-Contest.
  '''
  
  import sys
  from typing import Deque
  
  
  def input():
      return sys.stdin.readline().rstrip()
  
  
  DXY = [(0, -1), (1, 0), (0, 1), (-1, 0)]  # LDRU
  mod = 998244353
  inf = 1 << 64
  
  
  def ceil(x, y):
      return (x + y - 1) // y
  
  from collections import deque
  
  def slv():
      n, k, X = map(int, input().split())
      S = input()
      cnt = 0
      a = deque([])
      for i in range(n):
          if S[i] != '*':
              if cnt > 0:
                  a.append(k * cnt + 1)
              cnt = 0
          else:
              cnt += 1
          if i == n - 1 and cnt > 0:
              a.append(k * cnt + 1)
  
      total = 1
      b = deque([])
      while total <= X and a:
          v = a.pop()
          b.appendleft(v)
          total *= v
      a = deque([1]*(len(a)))
      c = deque([])
  
      for e in b:
          total //= e
          ord = ceil(X, total)
          c.append(ord)
          X -= (ord - 1) * total
  
      while c:
          a.append(c.popleft())
      string = []
  
      cnt, itr = 0, 0
      for i in range(len(S)):
          if S[i] == "a":
              string.append("a")
          else:
              if i + 1 == len(S) or S[i + 1] == "a":
                  string.append("b"*(a[itr] - 1))
                  itr += 1
      print("".join(string))
      return
  
  def main():
      t = int(input())
      for i in range(t):
          slv()
      return 0
    
  if __name__ == "__main__":
      main()