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;
  inline void debugMode() {}
  void solvecases(int cse) {
    int n, k;
    long long x;
    cin >> n >> k >> x;
    string s;
    cin >> s;
    vector<long long> d;
    int cnt = 0;
    for (int i = 0; i < n; ++i) {
      if (s[i] == '*')
        ++cnt;
      else {
        if (cnt) {
          d.push_back(cnt);
          cnt = 0;
        }
      }
    }
    if (cnt) d.push_back(cnt);
    if (d.empty()) {
      cout << s << '\n';
      return;
    }
    if (k == 0) {
      for (auto e : s) {
        if (e == 'a') cout << 'a';
      }
      cout << '\n';
      return;
    }
    for (auto &e : d) {
      e = e * k + 1;
    }
    vector<long long> ans((int)d.size(), 0);
    vector<long long> m((int)d.size() + 1);
    for (int i = 0; i < (int)d.size(); ++i) m[i] = LLONG_MAX;
    m[(int)d.size()] = 1;
    for (int i = (int)d.size() - 1; i >= 0; --i) {
      if (log10((long double)d[i]) + log10((long double)m[i + 1]) > 18.0) break;
      m[i] = m[i + 1] * d[i];
    }
    --x;
    for (int i = 1; i < (int)d.size(); ++i) {
      if (m[i] != LLONG_MAX && m[i] <= x) {
        ans[i - 1] = x / m[i];
        x %= m[i];
      }
    }
    ans[(int)d.size() - 1] = x;
    int i = 0, j = 0;
    string b;
    while (i < n) {
      if (s[i] == '*') {
        for (int kk = 0; kk < ans[j]; ++kk) b += 'b';
        ++j;
        while (i < n && s[i] == '*') {
          ++i;
        }
      } else {
        b += 'a';
        ++i;
      }
    }
    cout << b << '\n';
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    ;
    int t = 1;
    cin >> t;
    for (int i = 1; i <= t; i++) {
      solvecases(i);
    }
  }
  