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;
  void _print(long long t) { cerr << t; }
  void _print(int t) { cerr << t; }
  void _print(string t) { cerr << t; }
  void _print(char t) { cerr << t; }
  void _print(long double t) { cerr << t; }
  void _print(double t) { cerr << t; }
  void _print(unsigned long long t) { cerr << t; }
  template <class T, class V>
  void _print(pair<T, V> p);
  template <class T>
  void _print(vector<T> v);
  template <class T>
  void _print(set<T> v);
  template <class T, class V>
  void _print(map<T, V> v);
  template <class T>
  void _print(multiset<T> v);
  template <class T, class V>
  void _print(pair<T, V> p) {
    cerr << "{";
    _print(p.first);
    cerr << ",";
    _print(p.second);
    cerr << "}";
  }
  template <class T>
  void _print(vector<T> v) {
    cerr << "[ ";
    for (T i : v) {
      _print(i);
      cerr << " ";
    }
    cerr << "]";
  }
  template <class T>
  void _print(set<T> v) {
    cerr << "[ ";
    for (T i : v) {
      _print(i);
      cerr << " ";
    }
    cerr << "]";
  }
  template <class T>
  void _print(multiset<T> v) {
    cerr << "[ ";
    for (T i : v) {
      _print(i);
      cerr << " ";
    }
    cerr << "]";
  }
  template <class T, class V>
  void _print(map<T, V> v) {
    cerr << "[ ";
    for (auto i : v) {
      _print(i);
      cerr << " ";
    }
    cerr << "]";
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    int tc = 1;
    cin >> tc;
    while (tc--) {
      long long n, k, x;
      cin >> n >> k >> x;
      string second;
      cin >> second;
      vector<int> astericks;
      int cnt = 0;
      for (int i = n - 1; i >= 0; i--) {
        if (second[i] == '*') {
          cnt += k;
          if (i == 0) astericks.push_back(cnt);
        } else if (second[i] == 'a') {
          if (cnt) astericks.push_back(cnt);
          cnt = 0;
        }
      }
      if (x == 1) {
        for (int i = 0; i < n; i++)
          if (second[i] == 'a') cout << second[i];
        cout << "\n";
        continue;
      }
      vector<long long> dp(astericks.size(), LLONG_MAX);
      dp[0] = astericks[0] + 1;
      for (int i = 1; i < dp.size(); i++) {
        if (x / (astericks[i] + 1) < dp[i - 1]) break;
        dp[i] = (astericks[i] + 1) * dp[i - 1];
      }
      reverse(astericks.begin(), astericks.end());
      reverse(dp.begin(), dp.end());
      vector<int> ch(astericks.size());
      bool st = 0;
      for (int i = 0; i < ch.size(); i++) {
        if (x == dp[i]) {
          x = 0;
          for (int j = i; j < ch.size(); j++) ch[j] = astericks[j];
          break;
        } else if (x > dp[i]) {
          while (x > dp[i]) x -= dp[i], ch[i - 1]++;
          i--;
        } else if (i == ch.size() - 1)
          ch[ch.size() - 1] = x - 1;
      }
      int idx = 0;
      for (int i = 0; i < n; i++) {
        if (second[i] == '*') {
          while (i < n && second[i] == '*') i++;
          i--;
          for (int j = 0; j < ch[idx]; j++) cout << (char)('b');
          idx++;
        } else
          cout << (char)('a');
      }
      cout << "\n";
    }
  }
  