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;
const int inf = 1e9 + 7;
const long long big = inf * 1LL * inf;
const int maxn = 1e5 + 10;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int q;
  cin >> q;
  while (q--) {
    long long a, b, m;
    cin >> a >> b >> m;
    long long p2[51];
    p2[0] = 1;
    bool found = 0;
    for (int i = 1; i <= 50; ++i) p2[i] = p2[i - 1] * 2;
    for (int k = 1; k <= 50; ++k) {
      long long y = b - p2[max(k - 2, 0)] * (a + (k > 1));
      long long arr[k], exp2 = p2[max(k - 2, 0)];
      if (y < 0) break;
      for (int i = 1; i < k; ++i) {
        if (exp2 > 1) exp2 /= 2;
        arr[i] = min(y / exp2, m - 1);
        y -= exp2 * arr[i];
      }
      if (!y) {
        cout << k << ' ' << a << ' ';
        long long s = a;
        for (int i = 1; i < k; ++i) {
          cout << s + arr[i] + 1 << ' ';
          s += (s + arr[i] + 1);
        }
        found = 1;
        cout << '\n';
        break;
      }
    }
    if (!found) cout << -1 << '\n';
  }
  return 0;
}
