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;
const long long N = 1e2 + 10;
const long long mod = 1e9 + 7;
long long q;
long long a, b, m;
long long ans[N], sum, last;
vector<pair<long long, long long> > v;
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> q;
  for (long long o = 1; o <= q; ++o) {
    cin >> a >> b >> m;
    ans[1] = a;
    sum = a;
    last = 0;
    for (long long i = 2; i <= 50; ++i) {
      ans[i] = sum + 1;
      sum += ans[i];
      if (ans[i] > b) {
        last = i - 1;
        break;
      }
    }
    if (last == 0) {
      cout << -1 << endl;
      continue;
    }
    v.clear();
    long long temp = 0;
    for (long long i = 2; i < last; ++i) {
      ans[i] += min(m - 1, (b - ans[last]) / (long long)pow(2, last - i - 1));
      for (auto t : v) {
        ans[i] += ((long long)pow(2, i - t.second - 1)) * (t.first);
      }
      v.push_back(
          {min(m - 1, (b - ans[last]) / (long long)pow(2, last - i - 1)), i});
      ans[last] +=
          min(m - 1, (b - ans[last]) / (long long)pow(2, last - i - 1)) *
          (long long)pow(2, last - i - 1);
    }
    if (b - ans[last] > m - 1) {
      cout << -1 << endl;
      continue;
    }
    ans[last] = b;
    cout << last << ' ';
    for (long long i = 1; i <= last; ++i) {
      cout << ans[i] << ' ';
    }
    cout << endl;
  }
  return 0;
}
