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_base::sync_with_stdio(0);
  cin.tie(0);
  cout.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;
    }
    bool ok = 0;
    long long dif = 0;
    long long p = a, i = 1ll;
    while (p <= b) {
      if (b - p <= ((1ll << (i - 1ll)) * m) && b - p >= (1ll << (i - 1ll))) {
        ok = 1;
        dif = b - p - (1ll << (i - 1ll));
        break;
      }
      i++;
      p = (p * 2ll);
    }
    if (!ok) {
      cout << (-1) << '\n';
      continue;
    }
    long long s1 = min(dif, m - 1);
    dif -= s1;
    cout << i + 1 << ' ';
    long long res[55];
    res[0] = a;
    for (long long j = 1ll; j < i; j++) {
      long long sss = 0;
      for (long long u = 0; u < j; u++) sss += res[u];
      sss += min((dif / (1ll << (i - j - 1ll))), m - 1ll);
      dif -= min((dif / (1ll << (i - j - 1ll))), m - 1ll) *
             (long long)((1ll << (i - j - 1ll)));
      sss++;
      res[j] = sss;
    }
    long long sss = 0;
    for (long long u = 0; u < i; u++) sss += res[u];
    sss += s1;
    sss++;
    res[i] = sss;
    for (long long j = 0; j <= i; j++) cout << res[j] << ' ';
    cout << '\n';
  }
  return 0;
}
