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>
  #pragma GCC optimize("O3")
  #pragma GCC target("sse4")
  using namespace std;
  template <class T>
  using pq = priority_queue<T>;
  template <class T>
  using pqg = priority_queue<T, vector<T>, greater<T>>;
  template <class T>
  bool ckmin(T& a, const T& b) {
    return b < a ? a = b, 1 : 0;
  }
  template <class T>
  bool ckmax(T& a, const T& b) {
    return a < b ? a = b, 1 : 0;
  }
  mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
  const long long MOD = 1000000007;
  const char nl = '\n';
  const long long MX = 200001;
  void solve() {
    long long n, k, x;
    cin >> n >> k >> x;
    string s;
    cin >> s;
    vector<long long> v;
    long long cnt = 0;
    for (long long i = 0; i < s.size(); ++i) {
      if (s[i] != '*') {
        if (cnt) v.push_back(cnt * k);
        cnt = 0;
        continue;
      } else
        cnt++;
    }
    if (cnt) v.push_back(cnt * k);
    vector<long long> ans(v.size(), 0);
    while (!v.empty() && x != 0) {
      if (x <= v.back() + 1) {
        if (!v.empty()) ans[v.size() - 1] = min(x - 1, 1ll * v.back());
        break;
      }
      long long cur = v.back() + 1;
      for (long long i = v.size() - 2; i >= 0; --i) {
        if ((x - cur) / (v[i]) <= cur) {
          long long d = (x - cur) / (cur) + (((x - cur) % (cur)) > 0);
          ans[i] = d;
          x = x - (d * (cur));
          break;
        }
        cur += (v[i]) * (cur);
      }
    }
    long long idx = 0;
    for (long long i = 0; i < s.size(); ++i) {
      if (s[i] != '*') {
        if (i && s[i - 1] == '*') {
          while (ans[idx]--) cout << 'b';
          idx++;
        }
        cout << s[i];
      }
    }
    if (idx < v.size()) {
      while (ans[idx]--) cout << 'b';
      idx++;
    }
    cout << nl;
  }
  int main() {
    cin.tie(0)->sync_with_stdio(0);
    cin.exceptions(cin.failbit);
    long long T = 1;
    cin >> T;
    while (T--) {
      solve();
    }
    return 0;
  }
  