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 sys
  
  
  # sys.stdin = open('CP/input1.txt','r')
  # sys.stdout = open('CP/output1.txt','w')
  
  
  
  input=sys.stdin.readline
  
  def solve(n,k,x,string):
      x-=1
      newList=[]
      mul=[]
      temp=0
      for i in range(n):
          if(string[i]=='a'):
              if(temp>0):
                  newList.append(temp+1)
                  mul.append(temp+1)
                  temp=0
              newList.append(string[i])
          elif(string[i]=='*'):
              temp+=k
              if(i==n-1):
                  newList.append(temp+1)
                  mul.append(temp+1)
                  temp = 0
      mul.append(1)
      
      for i in range(len(mul)-2,0,-1):
          mul[i]=mul[i]*mul[i+1]
      mul=mul[1:]
      # print(mul,newList)
      ans=""
      lastIndex=0
      for i in range(len(newList)):
          if(newList[i]=='a'):
              ans+='a'
          else:
              t=((x//mul[lastIndex]))
  
              ans+=(t*'b')
              x-=t*mul[lastIndex]
              lastIndex+=1
      return ans
  
  
  
  t=int(input())
  for _ in range(t):
      n,k,x=map(int,input().split())
      string=input()
      print(solve(n,k,x,string))
  
  
  
  
  
  
      
      