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 heapq;import sys;import math;from collections import deque;from bisect import bisect_left,bisect_right;I=sys.stdin.readline;II=lambda :int(I());IN=lambda x:map(int,x.split());FN=lambda x:map(float,x.split());L=lambda x:list(IN(x));M=1000000007;P=print;T=True;F=False
  for _ in range(II()):
  	n,k,x=IN(I());x-=1
  	s=I()
  	# l=s[:-1].split('a')
  	#print(l)
  	st=[]#;v=1
  	i=0;lc=s[0];li=0
  	while(i<n):
  		while(s[i]==lc):
  			i+=1
  		st.append(i-li)
  		lc=s[i]
  		li=i
  	zx=len(st)-1
  	if(s[-2]=='a'):
  		zx-=1
  	wt=[1];f=-1
  	for i in range(zx,-1,-2):
  		twt=wt[-1]*(st[i]*k+1)
  		wt.append(twt)
  		if(x<twt):
  			f=i
  			break
  	res=[]#;print(wt,st,f)
  	i=0#;pos=0
  	if(s[0]=='a'):
  		i+=1
  		res.append('a'*st[0])
  	while(i<len(st)):
  		if(i>=f):
  			break
  		# pos+=1
  		i+=1
  		if(i<len(st)):
  			res.append('a'*st[i])
  		i+=1
  	while(i<len(st)):
  		q,x=divmod(x,wt[-2])
  		wt.pop()
  		res.append(q*'b')
  		i+=1
  		if(i<len(st)):
  			res.append('a'*st[i])
  		i+=1
  	print(*res,sep='')
  
  
  
  