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;
  inline void read(long long &x) {
    x = 0;
    long long f = 1;
    char ch = getchar();
    while (ch != 45 && (ch > '9' || ch < '0')) ch = getchar();
    if (ch == 45) {
      f = -1, ch = getchar();
    }
    while (ch <= '9' && ch >= '0') {
      x = x * 10 + ch - 48;
      ch = getchar();
    }
    x *= f;
  }
  void cmin(long long &a, long long b) { a = min(a, b); }
  void cmax(long long &a, long long b) { a = max(a, b); }
  const long long N = 2e3 + 10;
  long long i, j, k, n, s, t, m;
  char ch[N], ans[N];
  long long a[N], b[N];
  signed main() {
    read(t);
    while (t--) {
      read(n), read(k), read(s);
      m = 0;
      scanf("%s", ch + 1);
      memset(a, 0, sizeof(a));
      for (i = 1; i <= n; i++) {
        if (ch[i] != ch[i - 1]) ++m;
        if (ch[i] == 'a')
          a[m]--;
        else
          a[m]++;
      }
      s--;
      for (i = m; i >= 1; i--) {
        if (a[i] < 0) {
          b[i] = a[i];
          continue;
        }
        b[i] = s % (a[i] * k + 1);
        s /= (a[i] * k + 1);
      }
      for (i = 1; i <= m; i++) {
        for (j = 1; j <= abs(b[i]); j++) putchar((b[i] < 0 ? 'a' : 'b'));
      }
      cout << endl;
    }
    return 0;
  }
  