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;
inline void work() {
  long long a, b, m, r[100] = {0};
  cin >> a >> b >> m;
  if (a == b) {
    cout << "1 " << a << "\n";
    return;
  }
  for (long long n = 2;; ++n) {
    long long res = b - (a + 1 << n - 2);
    if (res < 0) break;
    if (res > (m - 1 << n - 2)) continue;
    for (int i = 2; i < n; ++i) {
      r[i] = min(res >> n - 1 - i, m - 1);
      res -= r[i] << n - 1 - i;
      ++r[i];
    }
    r[n] = res + 1;
    long long xi = a, si = a;
    cout << n << " " << xi;
    for (int i = 2; i <= n; ++i) {
      xi = si + r[i];
      si += xi;
      cout << " " << xi;
    }
    cout << "\n";
    return;
  }
  cout << "-1\n";
  ;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int q;
  cin >> q;
  while (q--) work();
  cout.flush();
  return 0;
}