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 math
  t = int(input())
  for case in range(t):
    n,k,x =list(map(int,input().split()))
    s = input()
    prod =[]
    i = n-1
    while i>=0:
      if s[i]=='*':
        cnt =1
        j =i-1
        while j>=0 and s[j]=='*':
          cnt+=1
          j-=1
        i=j
        
        if len(prod)==0:
          prod.append(cnt*k+1)
        else:
          prod.append((cnt*k+1)*prod[-1])
      else:
        i-=1
    
    prod =prod[::-1]
    
    m = len(prod)
    idx = 0
    ans =['']*(m)
    for i in range(m-1,-1,-1):
      if prod[i]>=x:
        idx =i
        break
    
    for i in range(idx ,m):
      if i==m-1:
        ans[i]='b'*(x-1)
        continue
      if prod[i]>=x:
        p = math.ceil(x/prod[i+1])-1
        ans[i]='b'*p
        x-=(p)*prod[i+1]
    i,j=0,0
    string =''
    while i<n:
      if s[i]=='*':
        string+=ans[j]
        j+=1
        while i<n and s[i]=='*':
          i+=1
      else:
        string+=s[i]
        i+=1
    print(string)
  
    
    
  
  
  