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;
  long long int gcd(long long int a, long long int b) {
    if ((a == 0) || (b == 0)) return a + b;
    return gcd(min(a, b), max(a, b) % min(a, b));
  }
  void solve() {
    long long int n, k, p;
    cin >> n >> k >> p;
    string s;
    cin >> s;
    long long int j;
    vector<long long int> temp;
    long long int cnt = 0;
    for (j = 0; j < n; j++) {
      if (s[j] == '*') {
        cnt++;
      } else if (cnt > 0) {
        temp.push_back(cnt * k + 1);
        cnt = 0;
      }
    }
    if (cnt > 0) {
      temp.push_back(cnt * k + 1);
    }
    long long int sz = temp.size();
    p--;
    for (j = sz - 1; j >= 0; j--) {
      if (p > 0) {
        long long int ok = temp[j];
        temp[j] = p % temp[j];
        p /= ok;
      } else {
        temp[j] = 0;
      }
    }
    sz = 0;
    string ans = "";
    for (j = 0; j < n; j++) {
      if (s[j] == 'a') {
        ans += 'a';
      } else {
        while (temp[sz] > 0) {
          ans += 'b';
          temp[sz]--;
        }
        sz++;
        while (j < n && s[j] == '*') {
          j++;
        }
        j--;
      }
    }
    cout << ans << "\n";
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    long long int t = 1;
    cin >> t;
    while (t--) {
      solve();
    }
    return 0;
  }
  