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 T;
long long a, b, m;
void work() {
  scanf("%lld%lld%lld", &a, &b, &m);
  if (a == b) {
    printf("%d %lld\n", 1, a);
    return;
  }
  if (a + m >= b && b >= a) {
    printf("%d %lld %lld\n", 2, a, b);
    return;
  }
  long long now = 1LL;
  int ans = 0;
  for (int i = 2; i <= 50; i++) {
    if (b / now >= a + 1 && (b + now - 1) / now <= a + m) {
      ans = i;
      break;
    }
    now = 2LL * now;
  }
  if (ans == 0)
    printf("%d\n", -1);
  else {
    printf("%d ", ans);
    long long cnt = b / now - a;
    long long r = b - 1LL * (cnt + a) * now;
    long long sum = a;
    printf("%lld ", a);
    now /= 2;
    for (int i = 1; i < ans; i++) {
      long long x = sum + cnt;
      if (r >= now && r != 0) {
        ++x;
        r -= now;
      }
      printf("%lld ", x);
      sum += x;
      now /= 2;
    }
    printf("\n");
  }
}
int main() {
  scanf("%d", &T);
  while (T--) work();
  return 0;
}
