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;
  int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    long long t, n, k, x;
    cin >> t;
    string st;
    while (t--) {
      cin >> n >> k >> x;
      x--;
      cin >> st;
      string s = "";
      vector<long long> v;
      for (long long i = 0; i < st.length();) {
        if (st[i] == 'a') {
          v.push_back(-1);
          i++;
        } else {
          long long c = 0;
          while (i < n && st[i] == '*') i++, c += k;
          v.push_back(c);
        }
      }
      vector<long long> v1(v.size());
      for (long long i = v1.size() - 1; i >= 0; i--) {
        if (v[i] < 0)
          v1[i] = -1;
        else {
          v1[i] = x % (v[i] + 1);
          x = x / (v[i] + 1);
        }
      }
      for (long long i = 0; i < v1.size(); i++) {
        if (v1[i] == 0) continue;
        if (v1[i] == -1)
          s += "a";
        else {
          for (long long j = 0; j < v1[i]; j++) s += "b";
        }
      }
      cout << s << endl;
    }
    return 0;
  }
  