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: sr=lambda: input()
  ir=lambda: int(sr())
  lr=lambda: list(map(int, sr().split()))
  
  inf = 10**18
  # mod = 10**9+7
  mod = 998244353
  
  if __name__=='__main__':
      test = ir()
      for t in range(test):
          n,k,x = lr()
          a=sr()
          count = []
          pre = ''
          now = 0
          start_inds = []
          for c in a:
              if c == '*':
                  if pre == '*':
                      now+=1
                  else:
                      now=1
              else:
                  if now > 0:
                      count.append(now*k+1)
                  now = 0
              pre = c
          if now>0 and pre == '*':
              count.append(now*k+1)
          count.reverse()
          count_per = [1]
          x-=1
          start = 0
          for ind, num in enumerate(count[:-1]):
              count_per.append(num*count_per[-1])
              if count_per[-1] > 10**18:
                  start = len(count)-1-ind
                  count_per.pop()
                  break
          for i in range(start):
              count.pop()
  
          count.reverse()
          count_per.reverse()
  
          nums = []
          ret = x
          for num in count_per:
              tmp = ret//num
              nums.append(tmp)
              ret = ret%num
  
          ans = []
          now_count = 0
          now = 0
          for c in a:
              if c=='*':
                  if pre=='*':
                      now+=1
                  else:
                      now=1
              else:
                  if now > 0:
                      if now_count >= start:
                          ind = now_count-start
                          for _ in range(nums[ind]):
                              ans.append('b')
                      now_count+=1
                  ans.append('a')
                  now = 0
              pre=c
          if pre=='*':
              if now_count >= start:
                  ind = now_count-start
                  for _ in range(nums[ind]):
                      ans.append('b')
              now_count+=1
          print(*ans, sep="")