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 long long mod = 998244353;
  const long long N = 1e5 + 10;
  template <typename... T>
  void debug(T&... args) {
    cout << "\n==================\n";
    ((cout << args << ", "), ...);
    cout << "\n==================\n";
  }
  template <typename T>
  void debug(vector<T>& a) {
    cout << "\n==============\n";
    for (auto& e : a) cout << e << ' ';
    cout << "\n==============\n";
  }
  template <typename... T>
  void put(T&... args) {
    ((cout << args << " "), ...);
  }
  void solve() {
    long long n, k, x;
    string s;
    cin >> n >> k >> x >> s;
    vector<long long> a;
    for (long long i = 0; i < n;) {
      if (s[i] == 'a') {
        a.push_back(-1);
        i++;
      } else {
        long long c = 0;
        while (i < n && s[i] == '*') i++, c += k;
        a.push_back(c);
      }
    }
    vector<long long> b(a.size() + 1);
    b[a.size()] = 1;
    x--;
    for (long long i = a.size() - 1; i > 0; i--) {
      long long prv = b[i + 1];
      if (a[i] < 0)
        b[i] = prv;
      else {
        if (a[i] + 1 > x / prv) {
          b[i] = x + 1;
        } else
          b[i] = prv * (a[i] + 1);
      }
    }
    string ans = "";
    for (long long i = 0; i < a.size(); i++) {
      if (a[i] < 0)
        ans += 'a';
      else {
        long long nxt = b[i + 1];
        while (x >= nxt) x -= nxt, ans += 'b';
      }
    }
    cout << ans << '\n';
  }
  signed main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    long long t;
    cin >> t;
    while (t--) solve();
  }
  