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: t = int(input())
  for T in range(t):
      n, k, x = map(int, input().split())
      s = input()
      x-=1
      cnt = 0
      v = []
      cntqq = 0
      for i in s:
          if i == '*':
              cnt+=1
              cntqq = 1
          else:
              if(cnt != 0):
                  v.append(cnt * k + 1)
                  cnt = 0
      if(cnt != 0):
          v.append(cnt * k + 1)
      p = [1] * len(v)
      if(cntqq):
          p[-1] = v[-1]
      for i in range(len(v) - 2, -1, -1):
          p[i] = p[i + 1] * v[i]
      
      p.append(0)
      flag = 1
      q = ''
      for i in range(len(s)):
          if(s[i] == 'a'):
              q += 'a'
          elif(len(q) == 0):
              q += '*'
          elif(q[-1] != '*'):
              q += '*'
      s = q
      #print(v)
      #print(s)
      #print(p)
      for i in range(len(s)):
          if(s[i] == 'a'):
              print('a', end='')
          elif(p[flag] == 0):
              print('b' * x, end='')
              x = 0
          elif(x == 0):
              x = 0
          else:
              while(x - p[flag] >= 0 and x > 0):
                  x -= p[flag]
                  print('b', end='')
              flag+=1
      print()
  