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;
signed main() {
  ios_base::sync_with_stdio(false);
  cout.tie(0);
  cin.tie(0);
  long long q;
  cin >> q;
  while (q--) {
    long long a, b, r;
    cin >> a >> b >> r;
    if (a == b) {
      cout << "1 " << a << "\n";
      continue;
    }
    long long tmp = -1;
    long long make;
    for (long long i = 2; i <= 50; i++) {
      long long sum = b - a * (pow(2, i - 2));
      if (sum <= 0) break;
      long long mx = r * pow(2, i - 2);
      long long mn = 1 * pow(2, i - 2);
      if (sum >= mn && sum <= mx) {
        tmp = i;
        make = sum;
        break;
      }
    }
    if (tmp == -1) {
      cout << "-1\n";
      continue;
    }
    vector<long long> ans;
    ans.push_back(a);
    vector<long long> br;
    long long k = tmp - 3;
    if (k == -1) k++;
    for (long long i = 0; i < tmp - 1; i++) {
      if (i == tmp - 2) {
        br.push_back(make);
        continue;
      }
      if (make - r * pow(2, k) >= pow(2, k)) {
        br.push_back(r);
        make -= r * pow(2, k);
      } else {
        br.push_back(1);
        make -= pow(2, k);
      }
      k--;
      if (k == -1) k++;
    }
    cout << tmp << " ";
    cout << a << " ";
    long long cum_sum = a;
    for (long long i = 0; i < tmp - 1; i++) {
      cout << cum_sum + br[i] << " ";
      cum_sum += cum_sum + br[i];
    }
    cout << "\n";
  }
}
