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 mod = 998244353;
  const long long MAX = 3e5 + 10;
  const long long Time = 86400;
  const long long X = 131;
  const long long inf = 0x3f3f3f3f;
  const double PI = 1e-4;
  double pai = 3.14159265358979323846;
  long long T, n, m, k, x, ans;
  string s;
  long long a[MAX];
  vector<long long> v;
  char ch;
  void solve() {
    v.clear();
    cin >> n >> k >> x;
    for (long long i = 1; i <= n; i++) {
      cin >> ch;
      if (ch == 'a')
        v.push_back(-1);
      else {
        if (v.empty() || v.back() == -1)
          v.push_back(k);
        else
          v.back() += k;
      }
    }
    a[v.size()] = 1;
    for (long long i = v.size() - 1; i >= 0; i--) {
      if (v[i] == -1) {
        a[i] = a[i + 1];
        continue;
      }
      if ((double)a[i + 1] * (v[i] + 1) > 2e18)
        a[i] = 2e18;
      else
        a[i] = a[i + 1] * (v[i] + 1);
    }
    for (long long i = 0; i < v.size(); i++) {
      if (v[i] == -1)
        cout << "a";
      else {
        while (x > a[i + 1]) cout << "b", x -= a[i + 1];
      }
    }
    cout << "\n";
  }
  signed main() {
    std::ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    cin >> T;
    while (T--) {
      solve();
    }
    return 0;
  }
  