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>
#pragma GCC optimize(2)
#pragma GCC optimize(3)
#pragma GCC optimize(4)
using namespace std;
int q;
long long a, b, m, p[55], d[55], sum[55];
int main() {
  scanf("%d", &q);
  sum[0] = p[0] = 1;
  for (int i = 1; i < 55; ++i) {
    if (i == 1)
      p[i] = 1, sum[i] = sum[i - 1] + p[i];
    else
      p[i] = p[i - 1] * 2, sum[i] = sum[i - 1] + p[i];
  }
  while (q--) {
    scanf("%lld %lld %lld", &a, &b, &m);
    bool f = false;
    if (a > b) {
      printf("-1\n");
      continue;
    }
    if (a == b) {
      printf("1 %lld\n", a);
      continue;
    }
    for (int i = 2; i <= 51; ++i) {
      long long l = a * p[i - 2], r = a * p[i - 2];
      long long tt = l;
      for (int j = i - 3; j >= 0; --j) {
        l += p[j];
        r += p[j] * m;
      }
      if (l > b) break;
      if (r < b) continue;
      for (int j = i - 3; j >= 0; --j) {
        if (j == 0) {
          if (b - tt <= m) d[0] = b - tt, f = true;
        } else {
          d[j] = (b - tt - sum[j - 1]) / p[j];
          d[j] = min(d[j], m);
          tt += d[j] * p[j];
        }
      }
      if (f) {
        vector<long long> vc;
        vc.push_back(a);
        long long s = a;
        for (int j = i - 3; j >= 0; --j) {
          long long t = s + d[j];
          s += t;
          vc.push_back(t);
        }
        printf("%d ", (int)vc.size());
        for (int i = 0; i < vc.size(); ++i)
          printf("%lld%c", vc[i], " \n"[i == vc.size() - 1]);
        break;
      }
    }
    if (!f) printf("-1\n");
  }
  return 0;
}
