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 MAXN = 55;
int q;
long long a, b, m, x[MAXN], y[MAXN], f[MAXN];
int main() {
  scanf("%d", &q);
  while (q--) {
    scanf("%I64d %I64d %I64d", &a, &b, &m);
    bool found = false;
    long long xb = 0, yb = 0;
    for (int k = 1; k <= 50; k++) {
      if (k == 1) {
        x[0] = a;
        y[0] = a;
        xb += a;
        yb += a;
      } else {
        x[k - 1] = 1;
        y[k - 1] = m;
        xb += 1;
        yb += m;
      }
      for (int i = 0; i < k - 2; i++) {
        xb += x[i];
        x[i] *= 2;
        yb += y[i];
        y[i] *= 2;
      }
      if (xb > b) break;
      if (yb >= b) {
        b -= xb;
        f[0] = a;
        for (int i = 1; i < k; i++) {
          long long d = min(m - 1, b / x[i]);
          f[i] = 1 + d;
          b -= x[i] * d;
        }
        printf("%d", k);
        long long fb = 0;
        for (int i = 0; i < k; i++) {
          if (i == 0) {
            fb += a;
          } else {
            fb += f[i];
          }
          for (int j = 0; j < i - 1; j++) {
            fb += f[j];
            f[j] *= 2;
          }
          printf(" %I64d", fb);
        }
        printf("\n");
        found = true;
        break;
      }
    }
    if (!found) printf("-1\n");
  }
  return 0;
}
