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;
  void _print(long long t) { cerr << t; }
  void _print(string t) { cerr << t; }
  void _print(char t) { cerr << t; }
  void _print(long double t) { cerr << t; }
  void _print(double t) { cerr << t; }
  void _print(unsigned long long t) { cerr << t; }
  template <class T, class V>
  void _print(pair<T, V> p);
  template <class T>
  void _print(vector<T> v);
  template <class T>
  void _print(set<T> v);
  template <class T, class V>
  void _print(map<T, V> v);
  template <class T>
  void _print(multiset<T> v);
  template <class T, class V>
  void _print(pair<T, V> p) {
    cerr << "{";
    _print(p.first);
    cerr << ",";
    _print(p.second);
    cerr << "}";
  }
  template <class T>
  void _print(vector<T> v) {
    cerr << "[ ";
    for (T i : v) {
      _print(i);
      cerr << " ";
    }
    cerr << "]";
  }
  template <class T>
  void _print(set<T> v) {
    cerr << "[ ";
    for (T i : v) {
      _print(i);
      cerr << " ";
    }
    cerr << "]";
  }
  template <class T>
  void _print(multiset<T> v) {
    cerr << "[ ";
    for (T i : v) {
      _print(i);
      cerr << " ";
    }
    cerr << "]";
  }
  template <class T, class V>
  void _print(map<T, V> v) {
    cerr << "[ ";
    for (auto i : v) {
      _print(i);
      cerr << " ";
    }
    cerr << "]";
  }
  template <typename T>
  istream &operator>>(istream &is, vector<T> &arr) {
    typename vector<T>::iterator a;
    for (a = arr.begin(); a != arr.end(); a++) {
      is >> *a;
    }
    return is;
  }
  template <typename T>
  ostream &operator<<(ostream &os, vector<T> &arr) {
    typename vector<T>::iterator a;
    for (a = arr.begin(); a != arr.end(); a++) {
      os << *a << " ";
    }
    return os;
  }
  template <typename T, typename U>
  ostream &operator<<(ostream &os, pair<T, U> &p) {
    os << p.first << " " << p.second << '\n';
    return os;
  }
  template <typename T, typename U>
  istream &operator>>(istream &is, pair<T, U> &p) {
    is >> p.first >> p.second;
    return is;
  }
  template <typename T1, typename T2>
  ostream &operator<<(ostream &os, vector<pair<T1, T2>> &arr) {
    typename vector<pair<T1, T2>>::iterator a;
    for (a = arr.begin(); a != arr.end(); a++) {
      os << *a;
    }
    return os;
  }
  template <typename T1, typename T2>
  istream &operator>>(istream &is, vector<pair<T1, T2>> &arr) {
    typename vector<pair<T1, T2>>::iterator a;
    for (a = arr.begin(); a != arr.end(); a++) {
      is >> *a;
    }
    return is;
  }
  bool isPrime(long long int n) {
    if (n <= 1) return false;
    if (n <= 3) return true;
    if (n % 2 == 0 || n % 3 == 0) return false;
    for (long long int i = 5; i * i <= n; i = i + 6)
      if (n % i == 0 || n % (i + 2) == 0) return false;
    return true;
  }
  vector<bool> isprime(2e5 + 10, true);
  vector<long long int> lp(2e5 + 10), hp(2e5 + 10);
  void sieve() {
    isprime[0] = isprime[1] = false;
    for (long long int i = 2; i < 2e5 + 10; i++) {
      if (isprime[i] == true) {
        for (long long int j = 2 * i; j < 2e5 + 10; j += i) {
          isprime[j] = false;
          hp[j] = i;
          if (lp[j] == 0) lp[j] = i;
        }
      }
    }
  }
  long long int add(long long int a, long long int b) {
    return ((a % 1000000007) + (b % 1000000007)) % 1000000007;
  }
  long long int sub(long long int a, long long int b) {
    return ((a % 1000000007) - (b % 1000000007) + 1000000007) % 1000000007;
  }
  long long int mul(long long int a, long long int b) {
    return ((a % 1000000007) * (b % 1000000007)) % 1000000007;
  }
  long long int mpow(long long int x, long long int y) {
    long long int res = 1;
    x = x % 1000000007;
    if (x == 0) return 0;
    while (y > 0) {
      if (y & 1) res = (res * x) % 1000000007;
      y = y >> 1;
      x = (x * x) % 1000000007;
    }
    return res;
  }
  void solve() {
    long long int n, k, x;
    cin >> n >> k >> x;
    string s;
    cin >> s;
    vector<long long int> v;
    long long int cnt = 0, acnt = 0;
    for (long long int i = 0; i < n; ++i) {
      if (s[i] == 'a') {
        v.push_back(-1);
        acnt++;
      } else {
        while (s[i] == '*' && i < n) {
          cnt++;
          i++;
        }
        v.push_back(cnt * k);
        cnt = 0;
        if (i != n) i--;
      }
    }
    n = v.size();
    string ans = "";
    if (x == 1) {
      for (long long int i = 0; i < acnt; ++i) ans += "a";
      cout << ans << "\n";
      return;
    }
    x--;
    for (long long int i = n - 1; i >= 0; i--) {
      if (v[i] == -1) {
        ans = "a" + ans;
      } else {
        string temp = "";
        long long int rem = x % (v[i] + 1);
        x = x / (v[i] + 1);
        for (long long int j = 0; j < rem; ++j) temp += "b";
        ans = temp + ans;
      }
    }
    cout << ans << "\n";
  }
  int32_t main() {
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    long long int t = 1;
    cin >> t;
    while (t--) {
      solve();
    }
    return 0;
  }
  