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;
long long x, y, z, n, m, q;
long long a[155];
string pat = "aoeui";
int main() {
  long long q;
  cin >> q;
  while (q--) {
    long long l, r, x;
    cin >> l >> r >> x;
    long long fl = 0;
    if (l == r) {
      cout << 1 << " " << l << endl;
      continue;
    }
    x--;
    for (int i = 0; i < 50; i++) {
      if ((l + 1) * (1LL << i) > r) {
        break;
      }
      long long lft = r - (l + 1) * (1LL << i);
      if (lft <= (1LL << i) * x) {
        a[0] = l;
        for (int j = 0; j <= i; j++) {
          a[j + 1] = 1;
          for (int k = 0; k <= j; k++) {
            a[j + 1] += a[k];
          }
        }
        for (int j = i; j >= 0; j--) {
          long long pr = j;
          if (pr > 0) {
            pr--;
          }
          long long px = (lft >> pr);
          px = min(px, x);
          lft -= px * (1LL << pr);
          for (int k = i + 1 - j; k <= i + 1; k++) {
            long long df = (k - (i + 1 - j));
            if (df > 0) {
              df--;
            }
            a[k] += px * (1LL << df);
          }
        }
        cout << i + 2;
        for (int k = 0; k <= i + 1; k++) {
          cout << " " << a[k];
        }
        cout << endl;
        fl = 1;
        break;
      }
    }
    if (!fl) {
      cout << -1 << endl;
    }
  }
  return 0;
}
