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 LIM = 2e18;
  long long n, k, x;
  string to_string(string s) { return '"' + s + '"'; }
  string to_string(const char* s) { return to_string((string)s); }
  string to_string(bool b) { return (b ? "true" : "false"); }
  template <typename A, typename B>
  string to_string(pair<A, B> p) {
    return "(" + to_string(p.first) + ", " + to_string(p.second) + ")";
  }
  template <typename A>
  string to_string(A v) {
    bool first = true;
    string res = "{";
    for (const auto& x : v) {
      if (!first) {
        res += ", ";
      }
      first = false;
      res += to_string(x);
    }
    res += "}";
    return res;
  }
  void debug_out() { cerr << '\n'; }
  template <typename Head, typename... Tail>
  void debug_out(Head H, Tail... T) {
    cerr << " " << to_string(H);
    debug_out(T...);
  }
  void solve() {
    cin >> n >> k >> x;
    x--;
    string s;
    cin >> s;
    long long cnt = count((s).begin(), (s).end(), '*');
    vector<int> cb;
    int c = 0;
    string t;
    for (int i = 0; i < n; i++) {
      if (s[i] == 'a') {
        t += 'a';
        if (c) {
          cb.push_back(c * k);
        }
        c = 0;
      } else {
        c++;
        if (c == 1) t += '*';
      }
    }
    if (c) {
      cb.push_back(c * k);
    }
    vector<int> bc;
    long long left = x;
    for (int i = 0; i < cb.size(); i++) {
      long long later = 1;
      for (int j = i + 1; j < cb.size(); j++) {
        if (later > LIM / (cb[j] + 1)) {
          later = LIM;
          break;
        }
        later *= cb[j] + 1;
      }
      42;
      bc.push_back(left / later);
      left %= later;
    }
    int h = 0;
    for (char c : t) {
      if (c == 'a')
        cout << c;
      else {
        while (bc[h]--) cout << 'b';
        h++;
      }
    }
    cout << '\n';
  }
  int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    int t;
    cin >> t;
    while (t--) {
      solve();
    }
    return 0;
  }
  