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 q, _c;
long long a, b, m, c[105];
int main() {
  for (scanf("%d", &q); q; q--) {
    _c = 0;
    memset(c, 0, sizeof(c));
    scanf("%I64d%I64d%I64d", &a, &b, &m);
    if (a == b) {
      printf("1 %I64d\n", a);
      continue;
    }
    long long l = a + 1, r = a + m;
    int n = 2;
    while (r < b) {
      l <<= 1;
      r <<= 1;
      n++;
    }
    if (l <= b && b <= r) {
      long long left = b - ((a + 1) << (n - 2)), sum = 0;
      for (int i = 2; i < n; i++) {
        if (sum + ((m - 1) << (n - i - 1)) < left) {
          sum += (m - 1) << (n - i - 1);
          c[i] = m - 1;
        } else {
          long long L = 0, R = m, mid;
          while (L < R - 1) {
            mid = (L + R) >> 1;
            if (sum + (mid << (n - i - 1)) <= left)
              L = mid;
            else
              R = mid;
          }
          sum += (L << (n - i - 1));
          c[i] = L;
        }
      }
      if (left > sum) c[n] = left - sum;
      printf("%d %I64d", n, a);
      sum = a;
      for (int i = 2; i <= n; i++) {
        printf(" %I64d", sum + 1 + c[i]);
        sum += sum + 1 + c[i];
      }
      puts("");
    } else {
      puts("-1");
    }
  }
  return 0;
}
