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;
  long long int n, k, x;
  string s;
  vector<int> vec, ans;
  inline void init() {}
  inline void input() {
    cin >> n >> k >> x >> s;
    x--;
  }
  inline void solve() {
    vec.clear();
    ans.clear();
    int cnt = 0;
    for (int i = 0; i < n; i++) {
      if (s[i] == '*') {
        cnt++;
      } else {
        vec.push_back(cnt * k);
        cnt = 0;
      }
    }
    vec.push_back(cnt * k);
    for (int i = vec.size() - 1; i > -1; i--) {
      ans.push_back(x % (vec[i] + 1));
      x /= (vec[i] + 1);
    }
    int sr = ans.size();
    for (int i = sr; i < vec.size(); i++) ans.push_back(0);
    string fin = "";
    reverse(ans.begin(), ans.end());
    for (int i = 0; i < ans.size(); i++) {
      string str = string(ans[i], 'b');
      fin = fin + str;
      if (i != ans.size() - 1) fin = fin + 'a';
    }
    cout << fin << endl;
  }
  int main() {
    ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
    int q = 1;
    init();
    cin >> q;
    while (q--) {
      input();
      solve();
    }
    return 0;
  }
  