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 t;
  scanf("%d", &t);
  while (t--) {
    long long int a, b, m, n, sum = 0, c = 0;
    scanf("%lld %lld %lld", &a, &b, &m);
    vector<long long int> arr;
    if (a == b) {
      printf("1 %lld\n", b);
      continue;
    }
    if (a + m >= b) {
      printf("2 %lld %lld\n", a, b);
      continue;
    }
    while (b > a) {
      arr.push_back(b);
      n = b;
      if (b % 2) c++;
      b = b / 2;
      if (c % 2) b += n % 2;
    }
    if (n - a > m)
      printf("-1\n");
    else {
      arr.push_back(a);
      auto it = arr.end() - 1;
      sum += (*it);
      while (1) {
        if (it == arr.begin()) break;
        it--;
        if ((*it) - sum > m) {
          printf("-1");
          sum = 0;
          break;
        }
        if ((*it) - sum == 0) {
          while (it < arr.end() - 1) {
            if ((*(it + 2)) + m >= (*it)) {
              sum -= (*(it + 1));
              arr.erase(it + 1);
              break;
            }
            it++;
          }
          if (it == arr.end() - 1) {
            printf("-1");
            sum = 0;
            break;
          }
        }
        sum += (*it);
      }
      if (sum) {
        it = arr.end() - 1;
        printf("%lld ", (long long int)arr.size());
        while (1) {
          printf("%lld ", (*it));
          if (it == arr.begin()) break;
          it--;
        }
      }
      printf("\n");
    }
  }
  return 0;
}
