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 sys import stdin, stdout, setrecursionlimit
  from math import gcd, sqrt, factorial, pi, inf
  from collections import deque, defaultdict
  from bisect import bisect, bisect_left
  from time import time
  from itertools import permutations as per
  from heapq import heapify, heappush, heappop, heappushpop
  
  input = stdin.readline
  R = lambda: map(int, input().split())
  I = lambda: int(input())
  S = lambda: input().rstrip('\r\n')
  L = lambda: list(R())
  P = lambda x: stdout.write(str(x) + '\n')
  lcm = lambda x, y: (x * y) // gcd(x, y)
  nCr = lambda x, y: (f[x] * inv((f[y] * f[x - y]) % N)) % N
  inv = lambda x: pow(x, N - 2, N)
  sumx = lambda x: (x ** 2 + x) // 2
  N = 10 ** 9 + 7
  
  for _ in range(I()):
      n, k, q = R()
      s = S()
      a = []
      cnt = []
      p = '#'
      for i in range(n):
          if s[i] == '*':
              if p == '*':
                  cnt[-1] += 1
              else:
                  a += '*',
                  cnt += 1,
          else:
              if p == 'a':
                  cnt[-1] += 1
              else:
                  a += 'a',
                  cnt += 1,
          p = s[i]
      v = [0] * len(a)
      p = 1
      sr = 1
      for i in range(len(a) - 1, -1, -1):
          if a[i] == '*':
              v[i] = p
              p = p * k * cnt[i] + sr
              sr = p
      ans = []
      q -= 1
      #print(v)
      for i in range(len(a)):
          if a[i] == 'a':
              ans += a[i] * cnt[i],
          else:
              if q >= v[i]:
                  ans += 'b' * (q // v[i]),
                  q %= v[i]
      print(*ans, sep='')
  