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() {
  int Q;
  cin >> Q;
  for (int q = 0; q < Q; q++) {
    long long a, b, m;
    cin >> a >> b >> m;
    long long n = floor(log2(b / a));
    long long rem = b - pow(2, n) * a;
    if (a == b) {
      cout << 1 << ' ' << a << endl;
      continue;
    }
    while (n >= 0 && (rem < pow(2, n) || rem > m * pow(2, n))) {
      n--;
      rem = b - pow(2, n) * a;
    }
    if (rem < pow(2, n) || rem > m * pow(2, n)) {
      cout << -1 << endl;
      continue;
    }
    long long val[1000];
    for (int i = 0; i < 1000; i++) val[i] = 0;
    val[n + 1] = a;
    for (int i = 0; i < n + 1; i++) val[i] = 1;
    rem -= pow(2, n);
    for (int i = n; i >= 0; i--) {
      long long add =
          min(m - 1, (long long)floor((double)rem / pow(2, max(i - 1, 0))));
      val[i] += add;
      rem -= add * pow(2, max(i - 1, 0));
    }
    long long sum = 0;
    cout << (n + 2) << ' ';
    for (int i = n + 1; i >= 0; i--) {
      cout << (sum + val[i]) << ' ';
      sum = 2 * sum + val[i];
    }
    cout << endl;
  }
}
