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;
  template <typename T, typename U>
  static inline void amn(T &x, U y) {
    if (y < x) x = y;
  }
  template <typename T, typename U>
  static inline void amx(T &x, U y) {
    if (x < y) x = y;
  }
  const long long MOD = 1e9 + 7;
  const long long INF = 2e18;
  const long long N = 5e5 + 5;
  long long n, k, x;
  string s, t;
  void solve() {
    cin >> n >> k >> x;
    cin >> s;
    t = "";
    vector<long long> v, vec;
    for (long long i = 0; i < n; i++) {
      t.push_back(s[i]);
      if (s[i] == '*') {
        long long j = i;
        while (j < n and s[j] == '*') {
          j++;
        }
        v.push_back(j - i);
        vec.push_back(0);
        i = j - 1;
      }
    }
    long long m = v.size();
    vector<long long> product(m + 1, 1);
    if (m > 0) product[m - 1] = v[m - 1] * k + 1;
    for (long long i = m - 2; i >= 0; i--) {
      long long p = (INF / product[i + 1]) + 1;
      if ((v[i] * k + 1) > p) {
        product[i] = INF;
      } else {
        product[i] = product[i + 1] * (v[i] * k + 1);
      }
      amn(product[i], INF);
    }
    for (long long i = 0; i < m; i++) {
      if (product[i + 1] >= x) {
        continue;
      }
      long long tmp = product[i + 1];
      vec[i] = (x - 1) / tmp;
      x -= vec[i] * tmp;
    }
    n = t.size();
    string ans;
    long long c = 0;
    for (long long i = 0; i < n; i++) {
      if (t[i] == 'a') {
        ans.push_back('a');
      } else {
        for (long long j = 1; j <= vec[c]; j++) {
          ans.push_back('b');
        }
        c++;
      }
    }
    cout << ans << "\n";
  }
  signed main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    long long test_case = 1;
    cin >> test_case;
    while (test_case--) {
      solve();
    }
    return 0;
  }
  