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;
  using ll = long long;
  constexpr int INF = 0x3f3f3f3f;
  constexpr long long LINF = 0x3f3f3f3f3f3f3f3fLL;
  constexpr double EPS = 1e-8;
  constexpr int MOD = 1000000007;
  constexpr int DY[]{1, 0, -1, 0}, DX[]{0, -1, 0, 1};
  constexpr int DY8[]{1, 1, 0, -1, -1, -1, 0, 1},
      DX8[]{0, -1, -1, -1, 0, 1, 1, 1};
  template <typename T, typename U>
  inline bool chmax(T& a, U b) {
    return a < b ? (a = b, true) : false;
  }
  template <typename T, typename U>
  inline bool chmin(T& a, U b) {
    return a > b ? (a = b, true) : false;
  }
  struct IOSetup {
    IOSetup() {
      std::cin.tie(nullptr);
      std::ios_base::sync_with_stdio(false);
      std::cout << fixed << setprecision(20);
    }
  } iosetup;
  template <typename T = char, typename U = std::string>
  std::vector<std::pair<T, int>> run_length_encoding(const U& s) {
    int n = s.size();
    std::vector<std::pair<T, int>> res;
    T now = s[0];
    int cnt = 1;
    for (int i = 1; i < n; ++i) {
      if (s[i] != now) {
        res.emplace_back(now, cnt);
        cnt = 0;
      }
      now = s[i];
      ++cnt;
    }
    res.emplace_back(now, cnt);
    return res;
  }
  void solve() {
    int n, k;
    ll x;
    cin >> n >> k >> x;
    string s;
    cin >> s;
    const auto rle = run_length_encoding(s);
    const int m = rle.size();
    string ans = "";
    vector<ll> w(m, 0);
    for (int i = m - 1; i >= 0; --i) {
      if (rle[i].first == '*') {
        const ll num = 1LL * k * rle[i].second + 1,
                 r = (i + 2 < m ? w[i + 2] : 1);
        w[i] = (r > LINF / num ? LINF : r * num);
      }
    }
    for (int i = (0); i < (m); ++i) {
      if (rle[i].first == 'a') {
        ans += string(rle[i].second, 'a');
        continue;
      }
      const ll r = (i + 2 < m ? w[i + 2] : 1);
      const ll put = (x - 1) / r;
      x -= put * r;
      ans += string(put, 'b');
    }
    assert(x == 1);
    cout << ans << '\n';
  }
  int main() {
    int t;
    cin >> t;
    while (t--) solve();
    return 0;
  }
  