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 i64 = long long;
void restore(const std::deque<i64> &v, i64 a) {
  int k = v.size();
  printf("%d %lld ", k + 1, a);
  i64 sum = a;
  for (int i = 0; i < k; i++) {
    printf("%lld ", sum + v[i] + 1);
    sum = (sum << 1ll) + v[i] + 1;
  }
  printf("\n");
}
int main() {
  int q;
  scanf("%d", &q);
  while (q--) {
    i64 a, b, m;
    scanf("%lld%lld%lld", &a, &b, &m);
    if (a == b) {
      printf("1 %lld\n", a);
      continue;
    }
    for (i64 k = 2; k <= 50; k++) {
      i64 rem = b - (1ll << (k - 2)) * (a + 1);
      if (rem < 0) break;
      if (rem > (1ll << (k - 2)) * (m - 1)) continue;
      std::deque<i64> v;
      for (i64 i = 2; i < k; i++) {
        i64 r = std::min(m - 1, rem / (1ll << (k - 1 - i)));
        v.push_back(r);
        rem -= r * (1ll << (k - 1 - i));
      }
      if (rem >= m) continue;
      v.push_back(rem);
      restore(v, a);
      goto next;
    }
    printf("-1\n");
  next:;
  }
  return 0;
}