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;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  long long q;
  cin >> q;
  while (q--) {
    long long a, b, m;
    cin >> a >> b >> m;
    if (a == b) {
      cout << "1 " << a << '\n';
      continue;
    }
    long long k;
    for (k = 1; k <= 50; k++) {
      if ((1ull << (k - 1)) * (a + 1) <= b && b <= (1ull << (k - 1)) * (a + m))
        break;
    }
    if (k > 50) {
      cout << "-1\n";
      continue;
    }
    vector<long long> r(k);
    for (long long i = 0; i < k; i++) r[i] = 1;
    b -= (1ull << (k - 1)) * (a + 1);
    for (long long i = 0; i < k; i++) {
      long long t = i == k - 1 ? 1 : (1ull << (k - i - 2));
      long long ta = min(b / t, m - 1);
      b -= ta * t;
      r[i] += ta;
    }
    if (b != 0) (0);
    vector<long long> res;
    res.push_back(a);
    for (long long i = 0; i < k; i++) {
      long long x = 0;
      for (long long j = 0; j < res.size(); j++) x += res[j];
      x += r[i];
      res.push_back(x);
    }
    cout << k + 1 << " ";
    for (long long i = 0; i < res.size(); i++) {
      cout << res[i];
      if (i == res.size() - 1)
        cout << '\n';
      else
        cout << ' ';
    }
  }
}
