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 q, m, k, a, b;
long long ans[200010], n;
long long base[200010];
inline long long read() {
  register long long w = 0, X = 0;
  register char ch = 0;
  while (!isdigit(ch)) {
    w |= ch == '-';
    ch = getchar();
  }
  while (isdigit(ch)) {
    X = (X << 1) + (X << 3) + (ch ^ 48);
    ch = getchar();
  }
  return w ? -X : X;
}
signed main() {
  base[0] = 1;
  for (register long long i = 1; i <= 50; i++) base[i] = base[i - 1] << 1;
  for (q = read(); q; q--) {
    a = read(), b = read(), m = read();
    n = 2;
    if (a == b) {
      printf("1 %lld\n", a);
      continue;
    }
    while ((a + m) * base[n - 2] < b) n++;
    if ((a + 1) * base[n - 2] <= b) {
      printf("%lld ", n);
      ans[1] = a, ans[n] = b;
      bool ifok = 0;
      for (register long long i = n - 1; i >= 2; i--) {
        if (b & 1) {
          b = (b + (ifok == 0 ? 1 : -1)) >> 1;
          ifok ^= 1;
        } else
          b >>= 1;
        ans[i] = b;
      }
      for (register long long i = 1; i <= n; i++) printf("%lld ", ans[i]);
      puts("");
    } else
      puts("-1");
  }
  return 0;
}
