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;
vector<long long> v;
int main() {
  int flag, T, i, j;
  long long a, b, m, tmp, res, del, now, dd;
  scanf("%d", &T);
  while (T--) {
    now = 1;
    flag = 0;
    scanf("%I64d%I64d%I64d", &a, &b, &m);
    if (a == b) {
      printf("%d ", 1);
      printf("%I64d\n", a);
      continue;
    }
    for (i = 1;; i++) {
      tmp = now * a;
      tmp += now;
      del = b - tmp;
      if (del < 0) {
        break;
      }
      dd = (m - 1) * (now);
      if (tmp + dd >= b) {
        del = b - tmp;
        v.clear();
        for (j = 1; j <= i; j++) {
          now = max(now / 2, 1LL);
          res = min(del / now, m - 1);
          del -= res * now;
          v.push_back(res);
        }
        flag = 1;
        printf("%d ", i + 1);
        printf("%lld ", a);
        long long sum = a;
        for (j = 1; j <= v.size(); j++) {
          printf("%I64d ", sum + v[j - 1] + 1);
          sum = sum * 2 + v[j - 1] + 1;
        }
        printf("\n");
        break;
      }
      now *= 2;
    }
    if (!flag) {
      printf("-1\n");
    }
  }
  return 0;
}
