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;
  vector<char> vp(4000005);
  void ans(vector<long long int> &dp, vector<long long int> &star,
           long long int x, vector<char> &vp, long long int n) {
    if (x <= 1) {
      return;
    }
    long long int starpos = upper_bound((dp).begin(), (dp).end(), x) - dp.begin();
    x -= dp[starpos - 1];
    if (x == 0) {
      for (long long int i = star[starpos - 1]; i < n; i++) {
        if (vp[i] == '*') {
          vp[i] = 'b';
        }
      }
    } else {
      for (long long int i = star[starpos]; i < n; i++) {
        while (i < n && vp[i] == '*') {
          vp[i] = 'b';
          i++;
        }
        break;
      }
      ans(dp, star, x, vp, n);
    }
  }
  int32_t main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    long long int T = 1;
    cin >> T;
    while (T--) {
      long long int n, k, x;
      cin >> n >> k >> x;
      string s;
      cin >> s;
      string b = "";
      for (long long int i = 0; i < k; i++) {
        b += 'b';
      }
      long long int cnt = 0;
      for (long long int i = 0; i < n; i++) {
        if (s[i] == '*') {
          long long int t = cnt;
          for (long long int j = cnt; j < t + k; j++) {
            vp[cnt] = '*';
            cnt++;
          }
        } else {
          vp[cnt] = 'a';
          cnt++;
        }
      }
      s.clear();
      vector<long long int> dp;
      vector<long long int> star;
      star.push_back(-1);
      dp.push_back(1);
      long long int prev = 1;
      long long int curr = 1;
      long long int flag1 = 1;
      for (long long int i = cnt - 1; i >= 0 && flag1; i--) {
        long long int flag = 0;
        while (i >= 0 && vp[i] == '*') {
          if (x < (curr + 1) * 1ll * prev) {
            dp.push_back((curr + 1) * 1ll * prev);
            star.push_back(i);
            flag1 = 0;
            break;
          }
          dp.push_back((curr + 1) * 1ll * prev);
          star.push_back(i);
          i--;
          curr++;
          flag = 1;
        }
        if (flag) {
          prev = curr * prev;
          curr = 1;
        }
      }
      long long int flag = 1;
      ans(dp, star, x, vp, cnt);
      for (long long int i = 0; i < cnt; i++) {
        if (vp[i] == '*') continue;
        cout << vp[i];
      }
      cout << "\n";
    }
    return 0;
  }
  