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;
long long arr[55];
long long r[55];
long long bin[55];
void init() {
  arr[0] = 0;
  arr[1] = 0;
  arr[2] = 1;
  for (int i = 3; i <= 50; ++i) {
    arr[i] = arr[i - 1] * 2;
  }
}
int main() {
  init();
  int n;
  cin >> n;
  while (n--) {
    long long a, b, c;
    cin >> a >> b >> c;
    if (a == b) {
      cout << 1 << " " << b << endl;
      continue;
    }
    int i;
    for (i = 2; i <= 50; ++i) {
      if (b - ((i - 1) * (a + 1) - a >= 0)) {
        long long add = b - a * arr[i];
        long long min_add = b / arr[i] - a;
        if (min_add <= c && min_add >= 1) {
          long long del = (b - a * arr[i]) - arr[i] * min_add;
          if (del > 0 && min_add + 1 > c) {
            continue;
          }
          memset(bin, 0, sizeof(bin));
          for (int j = i - 2; j > 0; --j) {
            bin[j] = del % 2;
            del /= 2;
          }
          r[0] = a;
          cout << i;
          cout << " " << a;
          for (int j = 1; j < i; ++j) {
            r[j] = 0;
            for (int k = 0; k < j; ++k) {
              r[j] += r[k];
            }
            if (i - j > del) {
              r[j] += min_add + bin[j];
            } else {
              r[j] += min_add + bin[j];
            }
            cout << " " << r[j];
          }
          cout << endl;
          break;
        }
      }
    }
    if (i > 50) {
      cout << "-1" << endl;
    }
  }
}
