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, k;
long long a, b, m, sum_min, sum_max, sum, cb, r;
long long Min[55], Max[55], p[55];
int main() {
  p[0] = 1;
  for (int i = 1; i <= 50; ++i) p[i] = p[i - 1] + p[i - 1];
  cin >> q;
  while (q--) {
    cin >> a >> b >> m;
    if (a == b) {
      cout << 1 << ' ' << a << '\n';
      continue;
    }
    Min[1] = a;
    Max[1] = a;
    sum_min = sum_max = a;
    k = -1;
    for (int i = 2; i <= 50; ++i) {
      Min[i] = sum_min + 1;
      Max[i] = sum_max + m;
      sum_min += Min[i];
      sum_max += Max[i];
      if (Min[i] <= b && Max[i] >= b) {
        k = i;
        break;
      } else if (Min[i] > b)
        break;
    }
    cout << k << ' ';
    if (k == -1) {
      cout << '\n';
      continue;
    }
    cb = b;
    cb -= p[k - 2] * a;
    for (int i = 2; i <= k; ++i) cb -= p[max(k - i - 1, 0)];
    cout << a << ' ';
    sum = a;
    --m;
    for (int i = 2; i <= k; ++i) {
      r = min(cb / p[max(k - i - 1, 0)], m);
      cb -= p[max(k - i - 1, 0)] * r;
      cout << sum + r + 1 << ' ';
      sum += sum + r + 1;
    }
    cout << '\n';
  }
  return 0;
}
