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::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  cout.precision(10);
  cout << fixed;
  unsigned long long q;
  cin >> q;
  while (q--) {
    unsigned long long a, b, m;
    cin >> a >> b >> m;
    if (a == b) {
      cout << 1 << " " << a << endl;
      continue;
    }
    if (b <= a + m) {
      if (b > a) {
        cout << "2 " << a << " " << b << endl;
      } else {
        cout << -1 << endl;
      }
      continue;
    }
    unsigned long long k = -1;
    for (unsigned long long i = 1; i <= 46; i++) {
      unsigned long long z0 = (1LL << i) * (a + 1);
      unsigned long long z1 = (1LL << i) * (a + m);
      if (b >= z0 && b <= z1) {
        k = i;
        break;
      }
    }
    if (k == -1)
      cout << "-1" << endl;
    else {
      unsigned long long pp = b - ((1LL << k) * (a + 1));
      unsigned long long aa = pp / ((1LL << k));
      unsigned long long ppp = pp % ((1LL << k));
      unsigned long long su = a;
      cout << (k + 2) << " " << a << " ";
      for (int i = 0; i < k + 1; i++) {
        unsigned long long ops = su + 1 + aa + ((ppp >> (k - i - 1)) & 1);
        cout << (ops) << " ";
        su += ops;
      }
      cout << endl;
    }
  }
  return 0;
}
