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;
  int dy[] = {1, -1, 0, 0, -1, 1, 1, -1};
  int dx[] = {0, 0, 1, -1, 1, -1, 1, -1};
  void file() {}
  void fast() {
    std::ios_base::sync_with_stdio(0);
    cin.tie(NULL);
  }
  long long mul(long long a, long long b) {
    if (a <= LLONG_MAX / b) return a * b;
    return (long long)1e18;
  }
  void test_case() {
    int n, k;
    long long x;
    cin >> n >> k >> x;
    string str;
    cin >> str;
    string tmp = "";
    vector<int> can;
    for (auto& ch : str) {
      if (ch == 'a') {
        can.push_back(0);
        tmp.push_back(ch);
        continue;
      }
      if (tmp.empty() || tmp.back() == 'a')
        tmp.push_back(ch), can.push_back(k);
      else
        can.back() += k;
    }
    str = tmp;
    tmp = "";
    n = (int)(str.size());
    vector<long long> suf(n + 1);
    suf.back() = 1;
    for (int i = n - 1; i >= 0; i--)
      suf[i] = (str[i] == 'a' ? suf[i + 1] : mul(can[i] + 1, suf[i + 1]));
    for (int i = 0; i < n; i++) {
      if (str[i] == 'a') {
        tmp.push_back('a');
        continue;
      }
      for (int j = 0; j <= can[i]; j++) {
        if (suf[i + 1] >= x) break;
        x -= suf[i + 1];
        tmp.push_back('b');
      }
    }
    cout << tmp << '\n';
  }
  int main() {
    file();
    fast();
    int t;
    cin >> t;
    while (t--) {
      test_case();
    }
    return 0;
  }
  