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 p[4000005];
  int sz[4000005];
  char s[4000005], t[4000005];
  int main() {
    long long x;
    int i, k, tt, n, m, cnt;
    scanf("%d", &tt);
    while (tt--) {
      cnt = 0;
      scanf("%d %d %lld %s", &n, &m, &x, s + 1);
      for (i = 1; i <= n; ++i)
        if (s[i] == 'a')
          t[++cnt] = 'a';
        else if (t[cnt] != '*') {
          t[++cnt] = '*';
          sz[cnt] = m;
        } else
          sz[cnt] += m;
      p[cnt + 1] = 1;
      for (i = n = cnt, cnt = 0; i; --i)
        if (t[i] == 'a')
          p[i] = p[i + 1];
        else {
          if (p[i + 1] == -1 || x / p[i + 1] < (sz[i] + 1))
            p[i] = -1;
          else
            p[i] = p[i + 1] * (sz[i] + 1);
        }
      for (i = 1; i <= n; ++i) {
        if (t[i] == 'a')
          s[++cnt] = 'a';
        else {
          if (p[i + 1] == -1)
            continue;
          else
            for (k = 1; k <= sz[i]; ++k)
              if (x > p[i + 1]) {
                s[++cnt] = 'b';
                x -= p[i + 1];
              }
        }
      }
      s[cnt + 1] = 0;
      puts(s + 1);
    }
    return 0;
  }
  