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 long long maxx = 1e14;
long long q, a, b, m;
long long f[50];
long long rr[59];
int main() {
  f[0] = 1;
  for (int i = 1; i <= 50; i++) f[i] = f[i - 1] * 2;
  cin >> q;
  while (q--) {
    cin >> a >> b >> m;
    if (b > maxx) {
      cout << -1 << endl;
      continue;
    }
    if (b == a) {
      cout << 1 << " " << a << endl;
      continue;
    } else {
      bool flag = 1;
      for (int k = 2; k <= 50; k++) {
        memset(rr, 0, sizeof(rr));
        long long t = b - f[k - 2] * a;
        long long ty = f[k - 2];
        if (t < 0) break;
        if (t >= ty && t <= ty * m) {
          long long tt = t / ty;
          long long y = t % ty;
          for (int i = 2; i <= k; i++) rr[i] = tt;
          for (int i = k - 3; i >= 0; i--)
            if (y >> i && (k - i - 1 >= 2)) rr[k - i - 1]++, y -= (1ll << i);
          cout << k << " " << a;
          for (int i = 2; i <= k; i++) {
            long long sum = rr[i];
            for (int j = 2; j <= i - 1; j++) sum = sum + rr[j] * f[i - j - 1];
            cout << " " << f[i - 2] * a + sum;
          }
          cout << endl;
          flag = 0;
          break;
        }
      }
      if (flag) cout << -1 << endl;
    }
  }
  return 0;
}
