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: from bisect import bisect_left, bisect_right
  import math
  
  for _ in range(int(input())):
      n,k,x = map(int,input().split())
      s = input()
  
      if(x == 1 or s.count('*') == 0):print('a'*(s.count('a'))); continue
  
      a = [1]; pos = []
      last = n-1
      while(last >=0 and s[last] == 'a'): last -=1
      c = 0
      while(last >=0 and s[last]== '*' ) : 
          last -=1; c+=1
      a.append(c*k +1)
      if(c!=0): pos.append([last+1,c*k])
  
      while(last>=0):
          while(last >=0 and s[last] == 'a'): last -=1
          c = 0
          if(last <0): break
          while(last >=0 and s[last]== '*' ) : 
              c += 1
              last -=1
          a.append(a[-1]*(k*c + 1))
          pos.append([last+1,k*c])
  
      ind = bisect_left(a,x)
      p = a[ind-1]
      
      d = math.ceil(x/p)
      left = d*p - x
      pos[ind-1][1] = d-1
  
      # print(left)
      for i in range(ind-2,-1,-1):
          if(left >= a[i+1]-a[i]):
              pos[i][1] = 0
              left -= a[i+1] - a[i]
          else:
              j = left//a[i]
              if(j == 0 and i == 0):
                  pos[i][1] -= left
                  left = 0
              else:
                  left %= a[i]
                  pos[i][1] -= j
              
          if(left== 0):break
  
      # print(a,ind,left,pos)
      i = 0; ans = ''; last = len(pos)
      while(i<n):
          while(i<n and s[i] == 'a'):
              ans += 'a'; i+=1
          if(i>=n): break
          while(i<n and s[i] == '*'):
              i += 1
          last -= 1
  
          if(last < ind):
              ans += 'b'*(pos[last][1])
      print(ans)