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;
long long int T, a, b, m, k, cnt;
signed main() {
  cout << fixed;
  cin >> T;
  while (T--) {
    cnt++;
    cin >> a >> b >> m;
    if (a == b) {
      cout << "1 " << a << endl;
      continue;
    }
    if (a + m >= b) {
      cout << "2 " << a << " " << b << endl;
      continue;
    }
    long long int k = ceil(log2(b) - log2(a + m) - 1e-9) + 2;
    if (k > 50) {
      cout << -1 << endl;
      continue;
    }
    long long int rem = (b) / (1LL << (k - 2)) - a, x = b % (1LL << (k - 2));
    if (rem < -1) {
      cout << "-1" << endl;
      continue;
    }
    long long int rvals[k + 2];
    for (long long int i = 2; i <= k; i++) rvals[i] = rem;
    bool canDo = true;
    for (long long int i = 2; i < k; i++) {
      rvals[i] += ((1LL << (k - 1 - i)) & x) > 0;
      if (rvals[i] <= 0 || rvals[i] > m) {
        cout << -1 << endl;
        canDo = false;
        break;
      }
    }
    if (!canDo) continue;
    long long int vals[k + 1], pre[k + 1];
    vals[1] = a;
    pre[1] = a;
    for (long long int i = 2; i <= k; i++) {
      vals[i] = pre[i - 1] + rvals[i];
      pre[i] = pre[i - 1] + vals[i];
    }
    if (vals[k] != b) {
      cout << -1 << endl;
      continue;
    }
    cout << k << " " << a;
    for (long long int i = 2; i <= k; i++) {
      cout << " " << vals[i];
    }
    cout << endl;
  }
}
