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 int maxn = 1e4 + 14;
int q;
long long a, b, m;
int main() {
  ios::sync_with_stdio(0), cin.tie(0);
  q = 1;
  cin >> q;
  while (q--) {
    cin >> a >> b >> m;
    vector<long long> v({a});
    bool done = 0;
    do {
      auto d = v;
      for (int i = 1; i < d.size(); i++) {
        long long lo = accumulate(d.begin(), d.begin() + i, 0ll), hi = lo + m;
        while (hi - lo > 1) {
          long long val = lo + hi >> 1;
          d[i] = val;
          bool ok = d[i] >= b;
          for (int j = i + 1; !ok && j < d.size(); j++)
            ok = (d[j] = accumulate(d.begin(), d.begin() + j, 0ll) + m) >= b;
          (ok ? hi : lo) = val;
        }
        d[i] = hi;
      }
      if (v.size() == 5) {
        for (auto x : d) cerr << x << ' ';
        cerr << '\n';
      }
      if (d.back() == b) {
        done = 1;
        cout << d.size() << ' ';
        for (auto x : d) cout << x << ' ';
        cout << '\n';
      }
      v.push_back(accumulate(v.begin(), v.end(), 0ll) + 1);
    } while (!done && v.back() <= b);
    if (!done) cout << "-1\n";
  }
}
