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 mul(long long int a, long long int b) {
    long long int ans = 0;
    while (b) {
      if (b & 1) {
        ans += a;
      }
      b = b >> 1;
      a += a;
      if (a > 1e18) {
        a = 1e18;
      }
      if (ans > 1e18) {
        ans = 1e18;
      }
    }
    return ans;
  }
  int32_t main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    ;
    long long int i, j, k, l, m, n, a, b, c, d, u, v, w, t = 1, x;
    cin >> t;
    while (t--) {
      string str;
      cin >> n >> k >> x;
      x--;
      cin >> str;
      vector<long long int> v1;
      long long int cnt = 0;
      for (long long int i = 0; i < n; i++) {
        if (str[i] == 'a')
          v1.push_back(cnt), cnt = 0;
        else
          cnt++;
      }
      v1.push_back(cnt);
      vector<long long int> v2(v1.size(), 1);
      for (long long int i = v1.size() - 2; i >= 0; i--) {
        v2[i] = mul(v2[i + 1], (k * v1[i + 1] + 1));
      }
      for (long long int i = 0; i < v2.size(); i++) {
        long long int u = x / v2[i];
        while (u) cout << 'b', u--;
        if (i + 1 != v1.size()) cout << 'a';
        x %= v2[i];
      }
      cout << endl;
    }
  }
  