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 int mod = 1e9 + 7;
  void fast() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
  }
  long long gcd(long long a, long long b) {
    if (b == 0) return a;
    return gcd(b, a % b);
  }
  long long lcm(long long x, long long y) { return x / gcd(x, y) * y; };
  long long fastpow(long long a, long long b) {
    long long wynik = 1;
    while (b) {
      if (b & 1) wynik = (wynik * a) % mod;
      a = (a * a) % mod;
      b >>= 1;
    }
    return wynik;
  }
  bool isp(long long a) {
    if (a == 2) return 1;
    if (a % 2 == 0 || a == 1) return false;
    for (long long i = 3; i * i <= a; i += 2) {
      if (a % i == 0) return false;
    }
    return true;
  }
  const int mx = 1e5 + 5;
  long long a[mx];
  long long fre[mx];
  bool sortPairs(const pair<char, int> &x, const pair<char, int> &y) {
    return (x.second > y.second);
  }
  int dx4[] = {-1, 0, 0, 1};
  int dy4[] = {0, 1, -1, 0};
  string dir[] = {"up", "rigth", "left", "down"};
  int dx8[] = {0, 0, 1, -1, 1, -1, 1, -1};
  int dy8[] = {1, -1, 0, 0, 1, -1, -1, 1};
  int pri[] = {-1, 0, 0, 0, 1, 1, 1, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0,
               1,  0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0,
               0,  0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0};
  int temp[] = {2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59};
  int arr22[4000];
  void solve() {
    unsigned long long t, n, k, x, arr1[2005], ha;
    string second, s1, ans;
    vector<long long> v;
    cin >> t;
    ha = t;
    while (t--) {
      cin >> n >> k >> x;
      cin >> second;
      for (int i = 0; i <= 2000; i++) {
        arr1[i] = 0;
      }
      for (int i = 0; i < n; i++) {
        long long num = 0;
        while (i < n && second[i] == '*') {
          i++;
          num++;
        }
        if (num) {
          v.push_back(num * k);
        }
      }
      arr1[v.size() - 1] = 1;
      for (int i = v.size() - 2; i >= 0; i--) {
        arr1[i] = (v[i + 1] + 1) * arr1[i + 1];
        if (arr1[i] > 1e18) {
          break;
        }
      }
      x--;
      int p = 0;
      for (int i = 0; i < n; i++) {
        if (second[i] == '*') {
          if (i > 0 && second[i - 1] == '*') continue;
          while (arr1[p] > 0 && arr1[p] <= x) {
            ans += 'b';
            x -= arr1[p];
          }
          p++;
        } else {
          ans += 'a';
        }
      }
      cout << ans << endl;
      ans = "";
      v.clear();
    }
  }
  int main() {
    fast();
    solve();
  }
  