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: from collections import defaultdict, deque, Counter
  from sys import stdin, stdout
  from heapq import heappush, heappop
  import math
  import io
  import os
  import math
  import bisect
  
  #?############################################################
  
  
  def isPrime(x):
      for i in range(2, x):
          if i*i > x:
              break
          if (x % i == 0):
              return False
      return True
  
  #?############################################################
  
  
  def ncr(n, r, p):
      num = den = 1
      for i in range(r):
          num = (num * (n - i)) % p
          den = (den * (i + 1)) % p
      return (num * pow(den, p - 2, p)) % p
  
  
  #?############################################################
  
  def primeFactors(n):
      l = []
      while n % 2 == 0:
          l.append(2)
          n = n / 2
      for i in range(3, int(math.sqrt(n))+1, 2):
          while n % i == 0:
              l.append(int(i))
              n = n / i
      if n > 2:
          l.append(n)
      return list(set(l))
  
  
  #?############################################################
  
  def power(x, y, p):
      res = 1
      x = x % p
      if (x == 0):
          return 0
      while (y > 0):
          if ((y & 1) == 1):
              res = (res * x) % p
          y = y >> 1
          x = (x * x) % p
      return res
  
  #?############################################################
  
  
  def sieve(n):
      prime = [True for i in range(n+1)]
      p = 2
      while (p * p <= n):
          if (prime[p] == True):
              for i in range(p * p, n+1, p):
                  prime[i] = False
          p += 1
      return prime
  
  
  #?############################################################
  
  def digits(n):
      c = 0
      while (n > 0):
          n //= 10
          c += 1
      return c
  
  #?############################################################
  
  
  def ceil(n, x):
      if (n % x == 0):
          return n//x
      return n//x+1
  
  #?############################################################
  
  
  def mapin():
      return map(int, input().split())
  
  #?############################################################
  
  
  # input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline
  # python3 15.py<in>op
  
  t = int(input())
  for _ in range(t):
      n, k, xi = mapin()
      s = list(input())
      ast = s.count('*')
      x = 0
      l = []
      for i in range(n):
          if s[i] == '*':
              x += 1
          else:
              l.append(x*k)
              x = 0
  
      l.append(x*k)
      ll =[0]*(len(l))
      # print(l)
      l.reverse()
      x = xi-1
      for i in range(len(l)):
          ll[i] = x%(l[i]+1)
          x = x//(l[i]+1)
      ll.reverse()
      ss = []
      for i in range(len(ll)):
          ss.append('b'*ll[i])
          ss.append('a')
      ss.pop(-1)
      print(''.join(ss))
      
  
  
      