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
  import os.path
   
  if(os.path.exists('input.txt')) :
      sys.stdin = open("input.txt", "r")
      sys.stdout = open("output.txt", "w")
      sys.stderr = open("error.txt", "w")
      
  depth = 200001
  mod = 1000000007 
  lim = mod * mod
  sys.setrecursionlimit(depth) 
   
  linp = lambda: list(minp())
  minp = lambda: map(int, input().split())
   
  from math import inf, ceil, sqrt, log2, gcd
  from collections import defaultdict, deque
  
  dxy = [(1, 0),(-1, 0),(0, 1),(0, -1)]
  
  def expo(a, b) :
      if not b : return 1
      v = expo(a, b//2)
      v = (v * v) % mod
      if b%2 : v = (v * a) % mod
      return v
  
  def modinv(a, b) :
      d = expo(b, mod-2)
      v = (a * d) % mod
      return v
  
  def checkPrime(n) :
      if n < 4 : return True
      if n%2 == 0 : return False
      for i in range(3, int(sqrt(n))+1, 2) :
          if n%i == 0 :
              return False
      return True    
  
  def highestPowerof2(x) :
      x |= x >> 1
      x |= x >> 2
      x |= x >> 4
      x |= x >> 8
      x |= x >> 16
  
      return x ^ (x >> 1)
  
  def isVowel(char):
      return char.lower() in 'aeiou'
  
  for _ in range(int(input())) :
      n, k, x = minp()
      s = str(input())
      a, c = list(), 0
      x = x - 1
  
      for e in s :
          if e == '*' :
              c = c + 1
          else :
              if c > 0 : a.append(c)
              c = 0
      if c > 0 : a.append(c)
  
      m = len(a)
      for i in range(m) :
          a[i] = a[i] * k + 1
  
      b = [a[i] for i in range(m)]
      for i in range(m-2, -1, -1) :
          b[i] = b[i] * b[i+1]
      b.append(1)
  
      c = list()
      for i in range(m) :
          v = x // b[i+1] 
          c.append(v)
          x = x - v * b[i+1]
  
      r, f, i = [], True, 0
      for e in s :    
          if e == "*" :
              f = False
          else :
              if not f :
                  r.append("b"*c[i])
                  i = i + 1
                  f = True
              r.append(e)
      if not f : 
          r.append("b"*c[i])
  
      print("".join(r))