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>
  ostream& operator<<(ostream& os, vector<T>& v) {
    if (v.size() == 0) {
      os << "empty vector\n";
      return os;
    }
    for (auto element : v) os << element << " ";
    return os;
  }
  template <typename T, typename second>
  ostream& operator<<(ostream& os, pair<T, second>& p) {
    os << "(" << p.first << ", " << p.second << ")";
    return os;
  }
  template <typename T>
  ostream& operator<<(ostream& os, set<T>& v) {
    if (v.size() == 0) {
      os << "empty set\n";
      return os;
    }
    auto endit = v.end();
    endit--;
    os << "[";
    for (auto it = v.begin(); it != v.end(); it++) {
      os << *it;
      if (it != endit) os << ", ";
    }
    os << "]";
    return os;
  }
  template <typename T>
  ostream& operator<<(ostream& os, multiset<T>& v) {
    if (v.size() == 0) {
      os << "empty multiset\n";
      return os;
    }
    auto endit = v.end();
    endit--;
    os << "[";
    for (auto it = v.begin(); it != v.end(); it++) {
      os << *it;
      if (it != endit) os << ", ";
    }
    os << "]";
    return os;
  }
  template <typename T, typename second>
  ostream& operator<<(ostream& os, map<T, second>& v) {
    if (v.size() == 0) {
      os << "empty map\n";
      return os;
    }
    auto endit = v.end();
    endit--;
    os << "{";
    for (auto it = v.begin(); it != v.end(); it++) {
      os << "(" << (*it).first << " : " << (*it).second << ")";
      if (it != endit) os << ", ";
    }
    os << "}";
    return os;
  }
  template <typename T>
  ostream& operator<<(ostream& os, vector<vector<T>>& v) {
    for (auto& subv : v) {
      for (auto& e : subv) os << e << " ";
      os << "\n";
    }
    return os;
  }
  bool do_debug = false;
  void Runtime_Terror() {
    long long n, k, tem;
    cin >> n >> k >> tem;
    long long x = tem;
    vector<string> v;
    string s;
    cin >> s;
    vector<long long> pos, maxi;
    for (long long i = 0; i < s.size(); i++) {
      if (s[i] == '*') {
        long long cnt = 0;
        pos.push_back(i);
        while (i < s.size() && s[i] == '*') {
          cnt++;
          i++;
        }
        maxi.push_back(cnt * k + 1);
      }
    }
    if (maxi.size() == 0) {
      cout << s << "\n";
      return;
    };
    ;
    vector<long long> ans(pos.size());
    vector<long long> pro(maxi.size());
    pro[pro.size() - 1] = maxi.back();
    long long sz = maxi.size();
    for (long long i = sz - 2; i >= 0; i--) {
      if (pro[i + 1] == -1) {
        pro[i] = -1;
        continue;
      }
      long double cal = (LLONG_MAX - 1e8) / (maxi[i] * 1.0);
      if (pro[i + 1] > (long long)(cal)) {
        pro[i] = -1;
        continue;
      }
      pro[i] = pro[i + 1] * maxi[i];
    }
    for (long long i = 0; i < maxi.size() - 1; i++) {
      if (pro[i + 1] == -1) {
        ans[i] = 1;
        continue;
      }
      long long small = pro[i + 1];
      for (long long j = 0; j < maxi[i] + 1; j++) {
        if (small >= x) {
          ans[i] = j + 1;
          break;
        }
        small += pro[i + 1];
      }
      x -= (ans[i] - 1) * (pro[i + 1]);
    }
    ans[ans.size() - 1] = x;
    ;
    long long in = 0;
    for (long long i = 0; i < s.size(); i++) {
      if (s[i] != '*')
        cout << s[i];
      else {
        for (long long j = 0; j < ans[in] - 1; j++) cout << "b";
        in++;
        while (i < s.size() && s[i] == '*') i++;
        i--;
      }
    }
    cout << "\n";
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    int t = 1;
    cin >> t;
    for (long long i = 0; i < t; i++) Runtime_Terror();
    return 0;
  }
  