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() {
  int TESTS = 1;
  while (TESTS--) {
    long long q;
    cin >> q;
    for (long long i = 0; i < q; i++) {
      long long a, b, m;
      cin >> a >> b >> m;
      if (a == b) {
        cout << 1 << ' ' << a << '\n';
      } else if (b - a <= m) {
        cout << 2 << ' ' << a << ' ' << b << '\n';
      } else {
        long long f[51];
        f[0] = 0;
        f[1] = 1;
        f[2] = 1;
        long long pre[51];
        pre[0] = 0;
        for (long long k = 3; k < 51; k++) f[k] = 2 * f[k - 1];
        for (long long k = 1; k < 51; k++) pre[k] = pre[k - 1] + f[k];
        long long ans = -1;
        long long x[51] = {0};
        for (long long k = 2; k < 50; k++) {
          if (ans != -1) break;
          for (long long j = 1; j < k + 1; j++) x[j] = 0;
          long long flag = 1;
          if (pre[k] + (pow(2, k - 1)) * a <= b) {
            long long z = b - (long long)pow(2, k - 1) * a;
            for (long long j = k + 1 - 1; j >= 2; j--) {
              long long lo = 0, h1 = m;
              while (lo < h1) {
                long long mid = (lo + h1 + 1) / 2;
                if (mid * f[j] + f[j] <= z)
                  lo = mid;
                else
                  h1 = mid - 1;
              }
              x[k - j + 1] = lo;
              if (!lo) {
                flag = 0;
                break;
              }
              z -= f[j] * lo;
            }
            if (!flag) continue;
            x[k] = z;
            if ((!x[k]) || (x[k] > m)) flag = 0;
            z = 0;
            if (flag) ans = k;
          } else
            break;
        }
        if (ans == -1) {
          cout << -1;
          cout << '\n';
        } else {
          cout << ans + 1 << ' ';
          long long tmp[51];
          tmp[0] = a;
          for (long long j = 1; j < ans + 1; j++) {
            tmp[j] = x[j];
            for (long long k = 0; k < j; k++) {
              tmp[j] += tmp[k];
            }
          }
          for (long long j = 0; j < ans + 1; j++) cout << tmp[j] << ' ';
          cout << '\n';
        }
      }
    }
  }
  return 0;
}
