Problem: 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.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int main() {
  int q, g;
  long long a, b, m, c, M, Dv[51], S, R[51], k;
  cin >> q;
  Dv[1] = 1;
  Dv[2] = 1;
  for (int i = 3; i <= 50; i++) {
    Dv[i] = Dv[i - 1] * 2;
  }
  for (int i = 0; i < q; i++) {
    g = 0;
    cin >> a >> b >> m;
    if (a == b) {
      cout << 1 << " " << a << "\n";
      g = 1;
    } else {
      for (int y = 2; y <= 50; y++) {
        S = Dv[y] * a;
        if (S > b) {
          break;
        }
        for (int u = 0; u < y - 1; u++) {
          R[y - u] = 1;
          S += Dv[y - u - 1];
        }
        if (S > b) {
          break;
        }
        for (int u = 0; u < y - 1; u++) {
          S -= Dv[y - u - 1];
          c = b - S;
          k = c / Dv[y - u - 1];
          k = min(k, m);
          R[y - u] = k;
          S += k * Dv[y - u - 1];
        }
        if (S == b) {
          g = 1;
          cout << y << " " << a << " ";
          c = a;
          S = a;
          for (int u = 0; u < y - 1; u++) {
            c = S + R[y - u];
            S += c;
            cout << c << " ";
          }
          cout << "\n";
          break;
        }
      }
    }
    if (g == 0) {
      cout << -1 << "\n";
    }
  }
  return 0;
}