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: #include <bits/stdc++.h>
  using namespace std;
  const int N = 1e6 + 10;
  const int mod = 7 + 1e9;
  inline int add(int a, int b) { return a + b >= mod ? a + b - mod : a + b; }
  inline int mul(int a, int b) { return a * 1ll * b % mod; }
  void wabalabadubdub() {
    long long n, k, x;
    string s, res;
    cin >> n >> k >> x >> s;
    --x;
    for (int i = s.size() - 1; i >= 0; --i) {
      if (s[i] == 'a') {
        res.push_back('a');
      } else {
        int j = i + 1;
        while (j > 0 && s[j - 1] == s[i]) --j;
        int b = x % ((i - j + 1) * k + 1);
        x /= ((i - j + 1) * k + 1);
        while (b--) res.push_back('b');
        i = j;
      }
    }
    reverse(res.begin(), res.end());
    cout << res << endl;
  }
  int main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    int t = 1;
    cin >> t;
    while (t--) wabalabadubdub();
  }
  