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 MX = 1e5 + 7;
vector<long long> v[MX];
bool vis[MX];
int main() {
  ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  long long t;
  cin >> t;
  while (t--) {
    long long a, b, m;
    cin >> a >> b >> m;
    if (a == b) {
      cout << 1 << " " << a << '\n';
      continue;
    }
    long long n = 0;
    for (long long k = 0; k < 48; k++) {
      if (((a + 1) << (k)) <= b && ((a + m) << (k)) >= b) {
        n = k + 2;
        break;
      }
      if (((a + m) << (k)) > b) break;
    }
    if (!n) {
      cout << -1 << '\n';
      continue;
    }
    long long c = b - a * (1ll << (n - 2)) - (1ll << (n - 2)), r[n];
    cout << n << " ";
    cout << a << " ";
    long long sum = a;
    for (long long k = 2; k < n; k++) {
      r[k] = min(m, c / (1ll << (n - k - 1)) + 1);
      c -= ((r[k] - 1) << (n - k - 1));
      cout << sum + r[k] << " ";
      sum += sum + r[k];
    }
    cout << b << '\n';
  }
}
