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 s[65], mx[65], mn[65], ans[65], mxx, mnn, anss, ss;
int main() {
  ss = 1ll;
  s[0] = 1ll;
  for (int i = 1; i < 53; ++i) s[i] = ss, ss += s[i];
  int n, cnt;
  long long a, b, m, shang;
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) {
    scanf("%I64d%I64d%I64d", &a, &b, &m);
    cnt = 1;
    mn[1] = a;
    mx[1] = a;
    mn[0] = mx[0] = 0;
    mxx = mnn = a;
    while (mx[cnt] < b) {
      ++cnt;
      mn[cnt] = 1ll + mnn;
      mx[cnt] = m + mxx;
      mnn += mn[cnt], mxx += mx[cnt];
    }
    if (b >= mn[cnt]) {
      --m;
      printf("%d ", cnt);
      b -= mn[cnt];
      for (int j = 1; j < cnt; ++j) {
        shang = b / s[cnt - j - 1];
        if (shang > m) shang = m;
        ans[j] = 1ll + shang;
        b -= (shang)*s[cnt - j - 1];
      }
      ans[0] = a;
      anss = a;
      for (int j = 0; j < cnt; ++j) {
        if (j > 0) ans[j] += anss, anss += ans[j];
        printf("%I64d ", ans[j]);
      }
      printf("\n");
    } else
      printf("-1\n");
  }
  return 0;
}
