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 int N = 20005;
  template <typename T, typename T1>
  T amax(T &a, T1 b) {
    if (b > a) a = b;
    return a;
  }
  template <typename T, typename T1>
  T amin(T &a, T1 b) {
    if (b < a) a = b;
    return a;
  }
  long long int solve() {
    long long int n, k, x;
    cin >> n >> k >> x;
    string s;
    cin >> s;
    vector<long long int> p;
    long long int cur = 0;
    string nev;
    for (long long int i = 0; i < n; ++i) {
      if (s[i] == 'a') {
        if (cur) nev.push_back('*');
        nev.push_back('a');
        if (cur) p.push_back(cur);
        cur = 0;
      } else {
        cur++;
      }
    }
    if (cur) nev.push_back('*');
    if (cur) p.push_back(cur);
    long long int sz = p.size();
    vector<long long int> ans(sz, 0);
    long long int pw = 1;
    long long int idx = sz - 1;
    x--;
    while (x) {
      cur = k * p[idx] + 1;
      ans[idx] = x % cur;
      x /= cur;
      idx--;
    }
    string y;
    long long int j = 0;
    for (long long int i = 0; i < nev.size(); ++i) {
      if (nev[i] == 'a')
        y.push_back('a');
      else {
        long long int m = ans[j];
        for (long long int l = 0; l < m; ++l) y.push_back('b');
        j++;
      }
    }
    return cout << y << "\n", 0;
  }
  signed main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    long long int test;
    cin >> test;
    while (test--) solve();
    return 0;
  }
  