Given a positive integer m, we say that a sequence x_1, x_2, ..., x_n of positive integers is m-cute if for every index i such that 2 ≤ i ≤ n it holds that x_i = x_{i - 1} + x_{i - 2} + ... + x_1 + r_i for some positive integer r_i satisfying 1 ≤ r_i ≤ m.

You will be given q queries consisting of three positive integers a, b and m. For each query you must determine whether or not there exists an m-cute sequence whose first term is a and whose last term is b. If such a sequence exists, you must additionally find an example of it.

Input

The first line contains an integer number q (1 ≤ q ≤ 10^3) — the number of queries.

Each of the following q lines contains three integers a, b, and m (1 ≤ a, b, m ≤ 10^{14}, a ≤ b), describing a single query.

Output

For each query, if no m-cute sequence whose first term is a and whose last term is b exists, print -1.

Otherwise print an integer k (1 ≤ k ≤ 50), followed by k integers x_1, x_2, ..., x_k (1 ≤ x_i ≤ 10^{14}). These integers must satisfy x_1 = a, x_k = b, and that the sequence x_1, x_2, ..., x_k is m-cute.

It can be shown that under the problem constraints, for each query either no m-cute sequence exists, or there exists one with at most 50 terms.

If there are multiple possible sequences, you may print any of them.

Example

Input


2
5 26 2
3 9 1


Output


4 5 6 13 26
-1

Note

Consider the sample. In the first query, the sequence 5, 6, 13, 26 is valid since 6 = 5 + \bf{\color{blue} 1}, 13 = 6 + 5 + {\bf\color{blue} 2} and 26 = 13 + 6 + 5 + {\bf\color{blue} 2} have the bold values all between 1 and 2, so the sequence is 2-cute. Other valid sequences, such as 5, 7, 13, 26 are also accepted.

In the second query, the only possible 1-cute sequence starting at 3 is 3, 4, 8, 16, ..., which does not contain 9.
#include <bits/stdc++.h>
using namespace std;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  unsigned long long tt;
  cin >> tt;
  while (tt--) {
    unsigned long long a, b, m;
    cin >> a >> b >> m;
    if (a == b) {
      cout << "1 " << a << '\n';
      continue;
    } else if (b - a <= m) {
      cout << "2 " << a << ' ' << b << '\n';
      continue;
    } else {
      unsigned long long i, j;
      unsigned long long l = 0;
      unsigned long long r = log2((unsigned long long)1e14) - 2;
      unsigned long long exp = 1;
      bool ok = 0;
      for (i = l; i <= r + 1; i++) {
        if (b >= exp * (a + 1) && b <= exp * (a + m)) {
          ok = 1;
          break;
        }
        exp *= 2;
      }
      if (ok) {
        vector<unsigned long long> r;
        r.push_back(b);
        bool f = 0;
        for (j = i; j > 0; j--) {
          unsigned long long t = r.back();
          if (t % 2) {
            if (!f)
              t -= t % 2;
            else
              t += t % 2;
            f = !f;
          }
          r.push_back(t / 2);
        }
        r.push_back(a);
        reverse(r.begin(), r.end());
        cout << i + 2 << ' ';
        for (auto x : r) {
          cout << x << ' ';
        }
        cout << '\n';
      } else {
        cout << "-1\n";
      }
    }
  }
}
