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, n;
unsigned long long int a, b, m;
unsigned long long int sequence[50], sum;
int possible(int i, unsigned long long int r) {
  unsigned long long int minEnd = (1LL << (n - i - 2)) * (2 * sum + r + 1);
  unsigned long long int maxEnd = (1LL << (n - i - 2)) * (2 * sum + r + m);
  if (minEnd > b) return (-1);
  if (minEnd <= b && b <= maxEnd) return (1);
  return (0);
}
int main() {
  scanf("%d", &q);
  while (q--) {
    scanf("%lld %lld %lld", &a, &b, &m);
    bool can = false;
    for (n = 2; n <= 50 && !can; n++)
      can = (1LL << (n - 2)) * (a + 1) <= b && b <= (1LL << (n - 2)) * (a + m);
    n--;
    if (a == b)
      printf("1 %lld\n", a);
    else if (!can)
      printf("-1\n");
    else {
      sequence[0] = sum = a;
      sequence[n - 1] = b;
      for (int j = 1; j < n - 1; j++) {
        unsigned long long int lo = 1, hi = m;
        while (lo < hi) {
          unsigned long long int mid = ((lo + hi) >> 1) + 0 * ((lo + hi) & 1);
          int verdict = possible(j, mid);
          if (verdict == -1)
            hi = mid - 1;
          else if (verdict == 1)
            hi = mid;
          else
            lo = mid + 1;
        }
        sequence[j] = sum + lo;
        sum += sequence[j];
      }
      printf("%d", n);
      for (int i = 0; i < n; i++) printf(" %lld", sequence[i]);
      printf("\n");
    }
  }
  return (0);
}
