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;
  void solve() {
    long long n, k, x;
    cin >> n >> k >> x;
    string s;
    cin >> s;
    s = '?' + s;
    vector<long long> v;
    long long len = 0;
    for (int i = n; i >= 1; i--) {
      if (s[i] != '*') {
        continue;
      }
      if (s[i] == '*' && s[i - 1] == '*') {
        len++;
      } else if (s[i] == '*') {
        len++;
        v.push_back(k * len + 1);
        len = 0;
      }
    }
    vector<long long> v1;
    for (int i = v.size() - 1; i >= 0; i--) {
      v1.push_back(v[i]);
    }
    long long ava = 1000000000000000000;
    if (v1.size() >= 2) {
      for (int i = v1.size() - 2; i >= 0; i--) {
        if ((ava / v1[i]) < v1[i + 1]) {
          for (int j = i; j >= 0; j--) {
            v1[j] = LLONG_MAX;
          }
          break;
        }
        v1[i] = v1[i] * v1[i + 1];
      }
    }
    int z = v1.size();
    v1.push_back(LLONG_MAX);
    vector<long long> ans;
    int i = 0;
    while (i < z) {
      if (i == z - 1) {
        ans.push_back(x);
        break;
      }
      long long q = ceil(x / (long double)v1[i + 1]);
      long long rem = x % v1[i + 1];
      x = rem;
      if (x == 0) {
        x = v1[i + 1];
      }
      if (i != z - 1) ans.push_back(q);
      i++;
    }
    int idx = 0;
    string t = "";
    for (int i = 1; i <= n; i++) {
      if (s[i] == 'a') {
        t += 'a';
        continue;
      }
      if (idx < ans.size()) {
        int req = ans[idx] - 1;
        for (int j = 0; j < req; j++) {
          t += 'b';
        }
        idx++;
      }
      for (int j = i; j <= n; j++) {
        if (s[j] == 'a') {
          i = j - 1;
          break;
        }
      }
    }
    cout << t << "\n";
  }
  int main() {
    ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
    int t = 1;
    cin >> t;
    while (t--) {
      solve();
    }
  }
  