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: # cook your dish here
  import sys
  #import random
  from bisect import bisect_left as lb
  from bisect import bisect_right as rb
  from collections import deque
  #sys.setrecursionlimit(10**8)
  from queue import PriorityQueue as pq
  from math import gcd
  #import math
  input_ = lambda: sys.stdin.readline().strip("\r\n")
  ii = lambda : int(input_())
  il = lambda : list(map(int, input_().split()))
  ilf = lambda : list(map(float, input_().split()))
  lii = lambda : list(map(int, list(ip())))
  ip = lambda : input_()
  fi = lambda : float(input_())
  ap = lambda ab,bc,cd : ab[bc].append(cd)
  li = lambda : list(input_())
  pr = lambda x : print(x)
  prinT = lambda x : print(x)
  f = lambda : sys.stdout.flush()
  inv =lambda x:pow(x,mod-2,mod)
  dx = [0,0,1,-1]
  dy = [1,-1,0,0]
  mod = 10**9 + 7
  mod1 = 998244353
  
  for _ in range (ii()) :
      n,k,x = il()
      s = list(ip())
  
      t = 0
  
      a = []
  
      for i in range (n) :
          if (s[i] == '*') :
              t += 1
          else :
              if (t != 0) :
                  a.append(['b',k * t + 1])
              t = 0
              a.append(['a',1])
  
      if (t != 0) :
          a.append(['b',k*t + 1])
  
      s = [1 for i in range (n+1)]
  
      for i in range (len(a) -1,-1,-1) :
          s[i] = s[i+1] * a[i][1]
          s[i] = min(10**19,s[i])
  
  
      ans = []
  
      for i in range (len(a)) :
          if (a[i][0] == 'b') :
              for j in range (a[i][1]) :
                  if (j+1)*s[i+1] >= x :
                      ans.append('b'*j)
                      x -= j * (s[i+1])
  
                      break
          else :
              ans.append(a[i][0])
  
          
          
      print("".join(ans))
  