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: def bas(n,k,x,s):
      i=0
      dc=[]
      cd=[]
      st=''
      while(i<len(s)):
          if s[i]=='*':
              j=i+1
              if j<len(s):
                  while(s[j]=='*'):
                      j=j+1
                      if j>=len(s):
                          break
              cd.append(j-i)
              i=j-1
          i=i+1
      i=0
      while(i<len(s)):
          if s[i]=='a':
              j=i+1
              if j<len(s):
                  while(s[j]=='a'):
                      j=j+1
                      if j>=len(s):
                          break
              dc.append(j-i)
              i=j-1
          i=i+1
      i=0
      while(i<len(cd)):
          cd[i]=cd[i]*k+1
          i=i+1
      i=0
      while(i<len(dc)):
          dc[i]=dc[i]*'a'
          i=i+1
      i=len(cd)-1
      p=1
      while(i>=0):
          if p>=x:
              break
          p=p*cd[i]
          i=i-1
      v=i+1
      i=len(cd)-1
      lans=[0]*len(cd)
      while(i>=v):
          lans[i]=x%cd[i]
          lans[i]=lans[i]-1
          if lans[i]<0:
              lans[i]=lans[i]+cd[i]
          if x%cd[i]!=0:
              x=x//cd[i]+1
          else:
              x=x//cd[i]
          i=i-1
      i=0
      while(i<len(lans)):
          if lans[i]==0:
              lans[i]=''
          else:
              lans[i]=lans[i]*'b'
          i=i+1
      ans=''
      if s[0]=='a':
          i=0
          j=0
          while(i<len(dc)):
              ans=ans+dc[i]
              if j<len(lans):
                  ans=ans+lans[j]
                  j=j+1
              i=i+1
      else:
          i=0
          j=0
          while(i<len(lans)):
              ans=ans+lans[i]
              if j<len(dc):
                  ans=ans+dc[j]
                  j=j+1
              i=i+1
      return ans
  t=int(input())
  l=[]
  for x in range(t):
      s1=str(input())
      s2=str(input())
      l.append([s1,s2])
  la=[]
  for x in l:
      lt=x[0].split(' ',)
      l1=[]
      for y in lt:
          l1.append(int(y))
      s=x[1]
      n=l1[0]
      k=l1[1]
      x=l1[2]
      ans=bas(n,k,x,s)
      la.append(ans)
  for x in la:
      print(x)