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 q;
long long a, b, m;
int main() {
  cin >> q;
  while (q--) {
    cin >> a >> b >> m;
    if (a == b) {
      cout << "1 " << a << "\n";
      continue;
    }
    for (int i = 2; i <= 51; ++i) {
      long long c = 1LL << (i - 2);
      if ((a + 1) * c > b) {
        cout << "-1\n";
        break;
      }
      long long d = b - a * c;
      if (d >= c && d <= m * c) {
        cout << i << " " << a;
        long long d = b - a * c;
        long long _a = a;
        long long _r = 0;
        for (int j = i - 2; j > 0; --j) {
          long long k = 1LL << j;
          long long r = min(m, 2 * d / k - 1);
          cout << " " << (_a + _r + r);
          _r *= 2;
          _r += r;
          _a *= 2;
          d = d - k * r / 2;
        }
        cout << " " << b << "\n";
        break;
      }
    }
  }
  return 0;
}
