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 n, k, x, tmp, f[2010], a[2010];
  string s;
  void solve() {
    cin >> n >> x >> k >> s;
    s = 'a' + s + 'a';
    memset(f, 0, sizeof f);
    f[n + 1] = 1;
    for (long long i = n; i >= 0; i--) {
      if (s[i] == 'a') {
        for (long long j = i + 1; j <= n + 1; j++) {
          if (s[j] == 'a') {
            if (f[j] != -1 && 1.0 * f[j] * ((j - i - 1) * x + 1) <= 1e18)
              f[i] = f[j] * ((j - i - 1) * x + 1);
            else
              f[i] = -1;
            break;
          }
        }
      }
    }
    for (long long i = 0; i <= n; i++) {
      if (s[i] == 'a') {
        if (i != 0) cout << 'a';
        for (long long j = i + 1; j <= n + 1; j++) {
          if (s[j] == 'a') {
            if (f[j] == -1) break;
            long long num = (k - 1) / f[j];
            for (long long o = 1; o <= num; o++) cout << 'b';
            k = (k - 1) % f[j] + 1;
            break;
          }
        }
      }
    }
    cout << '\n';
  }
  signed main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    long long _ = 1;
    cin >> _;
    while (_--) solve();
    return 0;
  }
  