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;
long long pot[100];
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> q;
  pot[0] = 1;
  for (int i = 1; i <= (60); i++) pot[i] = pot[i - 1] * 2LL;
  for (int I = 0; I < (q); ++I) {
    long long a, b, m;
    cin >> a >> b >> m;
    if (b == a)
      cout << 1 << " " << a << "\n";
    else {
      long long mozna = -1;
      for (long long i = 0; i <= 60; i++) {
        long long par = pot[i];
        if ((a + 1LL) * par > b) break;
        if ((a + 1LL) * par <= b && b <= (a + m) * par) {
          mozna = i;
          break;
        }
      }
      if (mozna == -1)
        cout << -1 << "\n";
      else {
        cout << mozna + 2 << " ";
        long long rem = b - a * pot[mozna];
        vector<long long> param;
        for (int i = mozna; i >= (0); i--) {
          long long ost;
          if (i) {
            ost = max(2LL, rem / pot[i - 1] - 1);
            while ((ost + 1) * pot[i - 1] <= rem) ost++;
          } else {
            ost = max(2LL, rem / pot[i] - 1);
            while (ost <= rem) ost++;
          }
          ost--;
          ost = min(ost, m);
          if (i) rem -= ost * pot[i - 1];
          param.push_back(ost);
        }
        long long sum = a;
        cout << sum << " ";
        for (auto x : param) {
          cout << sum + x << " ";
          sum += sum + x;
        }
        cout << "\n";
      }
    }
  }
  return 0;
}
