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 x) { cerr << x; }
  void __print(long x) { cerr << x; }
  void __print(unsigned x) { cerr << x; }
  void __print(unsigned long x) { cerr << x; }
  void __print(unsigned long long x) { cerr << x; }
  void __print(float x) { cerr << x; }
  void __print(double x) { cerr << x; }
  void __print(long double x) { cerr << x; }
  void __print(char x) { cerr << '\'' << x << '\''; }
  void __print(const char *x) { cerr << '\"' << x << '\"'; }
  void __print(const string &x) { cerr << '\"' << x << '\"'; }
  void __print(bool x) { cerr << (x ? "true" : "false"); }
  template <typename T, typename V>
  void __print(const pair<T, V> &x) {
    cerr << '{';
    __print(x.first);
    cerr << ',';
    __print(x.second);
    cerr << '}';
  }
  template <typename T>
  void __print(const T &x) {
    long long f = 0;
    cerr << '{';
    for (auto &i : x) cerr << (f++ ? "," : ""), __print(i);
    cerr << "}";
  }
  void _print() { cerr << "]\n"; }
  template <typename T, typename... V>
  void _print(T t, V... v) {
    __print(t);
    if (sizeof...(v)) cerr << ", ";
    _print(v...);
  }
  vector<long long> factorise(long long x) {
    vector<long long> v;
    for (long long i = 1; i * i <= x; i++) {
      if (x % i == 0) {
        v.push_back(i);
        if (i * i != x) v.push_back(x / i);
      }
    }
    return v;
  }
  long long power(long long x, long long y, long long p) {
    long long res = 1;
    x %= p;
    if (x == 0) return 0;
    while (y > 0) {
      if (y & 1) res = (res * x) % p;
      y = y >> 1;
      x = (x * x) % p;
    }
    return res;
  }
  long long nCk(long long x, long long k) {
    long long num = 1;
    for (long long i = 0; i < k; i++) {
      num *= (x - i) % 1000000007;
      num %= 1000000007;
    }
    long long den = 1;
    for (long long i = 1; i <= k; i++) {
      den *= i % 1000000007;
      den %= 1000000007;
    }
    return (num * power(den, 1000000007 - 2, 1000000007)) % 1000000007;
  }
  template <typename T>
  bool is_palindrome(T v, long long n) {
    bool bl = true;
    for (long long i = 0; i < n / 2; i++) {
      if (v[i] != v[n - i - 1]) {
        bl = false;
        break;
      }
    }
    return bl;
  }
  bool cmp(vector<long long> &a, vector<long long> &b) {
    if (a[0] < b[0])
      return true;
    else if (a[0] > b[0])
      return false;
    else {
      if (a[1] - a[0] < b[1] - b[0]) return true;
      return false;
    }
    return false;
  }
  signed main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    long long test;
    cin >> test;
    long long te = 0;
    while (test--) {
      te++;
      long long n, k, x;
      cin >> n >> k >> x;
      vector<long long> v;
      v.push_back(1);
      string s;
      cin >> s;
      long long t = 0;
      for (long long i = n - 1; i >= 0; i--) {
        if (s[i] == '*') {
          t++;
        } else if (t) {
          if (v[v.size() - 1] > x / (t * k + 1))
            v.push_back(LLONG_MAX);
          else
            v.push_back(1ll * (t * k + 1) * v[v.size() - 1]);
          t = 0;
        }
      }
      if (t) {
        if (v[v.size() - 1] > x / (t * k + 1))
          v.push_back(LLONG_MAX);
        else
          v.push_back(1ll * (t * k + 1) * v[v.size() - 1]);
      };
      long long i = v.size() - 2;
      long long j = 0;
      x--;
      while (j < s.length()) {
        while (j < s.length() && s[j] != '*') {
          cout << s[j];
          j++;
        }
        if (j < s.length() && s[j] == '*') {
          long long temp = x / v[i];
          x = x % v[i];
          ;
          cout << string(temp, 'b');
          i--;
        }
        while (j < s.length() && s[j] == '*') j++;
      }
      cout << "\n";
    }
    return 0;
  }
  