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 N = 3500;
  char s[N];
  unsigned long long d[N];
  unsigned long long id[N];
  unsigned long long tx[N];
  int main() {
    int T;
    cin >> T;
    while (T--) {
      memset(tx, 0, sizeof tx);
      memset(d, 0, sizeof d);
      memset(id, 0, sizeof id);
      int n, k;
      cin >> n >> k;
      unsigned long long x;
      cin >> x;
      cin >> s;
      x--;
      bool falg = false;
      if (k == 0) {
        for (int i = 0; i < n; i++) {
          if (s[i] == 'a') cout << s[i];
        }
        cout << endl;
        continue;
      }
      unsigned long long cnt = 0;
      unsigned long long j = 0;
      for (int i = 0; i <= n - 1; i++) {
        if (s[i] == 'a' && cnt != 0) {
          id[j] = i;
          d[j++] = cnt;
          cnt = 0;
        } else if (s[i] == '*') {
          cnt += k;
        }
      }
      if (cnt) id[j] = n, d[j++] = cnt;
      unsigned long long last = 1;
      int m = -1;
      for (int i = j - 1; i >= 0; i--) {
        if (d[i] * last >= x) {
          m = i;
          tx[i] = x / last;
          x -= last * tx[i];
          break;
        } else
          last = (last) * (d[i] + 1);
      }
      for (int i = m + 1; i < j; i++) {
        if (!x) break;
        last /= (d[i] + 1);
        tx[i] = x / last;
        x -= last * tx[i];
      }
      int tt = 0;
      for (int i = 0; i < n; i++) {
        if (s[i] == 'a') {
          cout << "a";
        } else {
          for (int z = 0; z < tx[tt]; z++) cout << "b";
          i = id[tt] - 1;
          tt++;
        }
      }
      cout << endl;
    }
  }
  