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 math
  from collections import deque,Counter
  from sys import stdin
  
  #sys.setrecursionlimit(10**7)
  
  int1=lambda x: int(x)-1
  stinput=lambda :stdin.readline()[:-1]
  ii=lambda :int(stinput())
  mi=lambda :map(int, stdin.readline().split())
  li=lambda :list(mi())
  mi1=lambda :map(int1, stdin.readline().split())
  li1=lambda :list(mi1())
  mis=lambda :map(str, stdin.readline().split())
  lis=lambda :list(mis())
  
  from collections import defaultdict
  #d=defaultdict(int)
  #d=defaultdict(lambda:1)
  
  import heapq
  #heapq.heapify(hq)
  #heapq.heappop(hq)
  #heapq.heappush(hq,1)
  
  mod=10**9+7
  Mod=998244353
  INF=10**18
  ans=0
  
  t=ii()
  for _ in range(t):
    a,b,c=mi()
    s=stinput()    
    x=[]
    cnt=0
    ans=[]
    if c==1:
      for i in s:
        if i=='a':
          ans.append(i)
      print(''.join(ans))
      continue
    for i in s:
      if i=='a':
        if cnt>0:
          x.append(cnt+1)
          ans.append(-1)
        cnt=0
        ans.append('a')
      else:
        cnt+=b
    if cnt>0:
      ans.append(-1)
      x.append(cnt+1)
    tmp=1
    d=1
    c-=1
    #print(x)
    while c!=0:
      x[-d],c=c%x[-d],c//x[-d]    
      d+=1
    for i in range(len(x)-d+1):
      x[i]=0
    tmp=0
    #print(x)
    for i in range(len(ans)):
      if ans[i]==-1:
        ans[i]='b'*x[tmp]
        tmp+=1
    print(''.join(ans))
  