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 main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int t;
  cin >> t;
  while (t--) {
    long long int a, b, m;
    cin >> a >> b >> m;
    if (a == b) {
      cout << "1 " << a << "\n";
      continue;
    }
    int yes = 0;
    int tr;
    vector<long long int> ans;
    for (int i = 2; 1; i++) {
      long long int cur = (1LL << (i - 2));
      if (cur * (a + 1) > b) break;
      long long int ch = cur * (a + m);
      if (ch < b) continue;
      tr = i;
      yes = 1;
      long long int num = b / cur;
      long long int rem = b % cur;
      num -= a;
      for (int j = 1; j <= i - 1; j++) ans.push_back(num);
      ans.push_back(a);
      for (int j = 0; j <= 49; j++) {
        if (rem & (1LL << j)) ans[j + 1] += 1;
      }
      break;
    }
    if (!yes)
      cout << "-1\n";
    else {
      cout << tr << " ";
      cout << a << " ";
      for (int i = 2; i <= tr; i++) {
        long long int disp = 0;
        long long int cur = (1LL << (i - 2));
        int p = ans.size() - 1;
        while (cur) {
          disp += cur * ans[p];
          cur /= 2, p--;
        }
        disp += ans[p];
        cout << disp << " ";
      }
      cout << "\n";
    }
  }
  return 0;
}
