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;
const int iinf = 0x3f3f3f3f;
const long long linf = 0x3f3f3f3f3f3f3f3f;
const int N = 50;
long long a, b, m, ans[N];
void Main() {
  cin >> a >> b >> m, --m;
  if (a == b) return void(cout << 1 << " " << a << '\n');
  for (int len = (2), I = (N + 1); len < I; len++) {
    long long lst = b - (a << (len - 2)) - ((1ll << (len - 2)) - 1) - 1,
              sum = 0;
    if (lst < 0) return void(cout << -1 << '\n');
    for (int i = (1), I = (len - 1); i < I; i++)
      ans[i] = min(m, lst / (1ll << (len - i - 2))),
      lst -= (ans[i] << (len - i - 2));
    if (lst <= m) {
      cout << len << " " << a << " ", sum += a;
      for (int i = (1), I = (len - 1); i < I; i++)
        cout << (sum + ans[i] + 1) << " ", sum += sum + ans[i] + 1;
      return void(cout << b << '\n');
    }
  }
  cout << -1 << '\n';
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0), cout.tie(0);
  int ta;
  cin >> ta;
  while (ta--) Main();
  return 0;
}
