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;
  string s;
  int n, k, t, cnt[2001];
  unsigned long long x;
  int main() {
    cin >> t;
    while (t--) {
      cin >> n >> k >> x;
      cin >> s;
      string news = "";
      int len = -1;
      for (int i = 0; i < n; i++) {
        if (i == 0) {
          news += s[i];
          if (s[i] == '*') {
            len++;
            cnt[len] = k;
          } else {
            len++;
            cnt[len] = 0;
          }
        } else {
          if (s[i] == s[i - 1] && s[i] == '*') {
            cnt[len] += k;
          } else {
            news += s[i];
            if (s[i] == '*') {
              len++;
              cnt[len] = k;
            } else {
              len++;
              cnt[len] = 0;
            }
          }
        }
      }
      len = news.size();
      unsigned long long tt = 1, vt = 0;
      vector<unsigned long long> thutu;
      thutu.push_back(1);
      for (int i = len - 1; i >= 0; i--) {
        unsigned long long prett = tt;
        if (cnt[i] > 0) {
          tt = tt * (cnt[i] + 1);
          thutu.push_back(tt);
          if (x / prett <= cnt[i]) {
            vt = i;
            break;
          }
        }
      }
      for (int i = 0; i < vt; i++) {
        if (news[i] == 'a') {
          cout << news[i];
        }
      }
      if (thutu.size()) {
        thutu.pop_back();
      }
      for (int i = vt; i < len; i++) {
        if (news[i] == 'a') {
          cout << news[i];
        } else {
          if (thutu.size()) {
            unsigned long long giasu = thutu.back();
            thutu.pop_back();
            unsigned long long o = x / giasu;
            if (x % giasu == 0) {
              o--;
            }
            x -= o * giasu;
            for (int j = 1; j <= o; j++) {
              cout << 'b';
            }
          }
        }
      }
      cout << "\n";
    }
  }
  