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 = 2e5 + 5;
unsigned long long arrmin[maxn];
int main() {
  int q;
  cin >> q;
  for (int i = 0; i < q; i++) {
    unsigned long long a, b, mod;
    cin >> a >> b >> mod;
    if (b - a <= mod) {
      if (a == b) {
        cout << "1 " << a << endl;
      } else
        cout << "2 " << a << " " << b << endl;
      continue;
    }
    unsigned long long arrmax;
    arrmin[0] = a;
    arrmin[1] = a + 1;
    arrmax = a + mod;
    int num = 2;
    for (num; 2 * arrmin[num - 1] <= b; num++) {
      arrmin[num] = 2 * arrmin[num - 1];
      arrmax = 2 * arrmax;
    }
    if (arrmax >= b || arrmax < arrmin[num - 1]) {
      cout << num << " ";
      for (int i = 1; i < num - 1; i++) {
        unsigned long long l = arrmin[i];
        unsigned long long r = arrmin[i] + mod - 1;
        unsigned long long base = 1;
        while (l < r) {
          unsigned long long m = (l + r) >> 1;
          unsigned long long tmp = (m - arrmin[i]) * (base << num - i - 2);
          if (b >= arrmin[num - 1] + tmp)
            l = m + 1;
          else
            r = m - 1;
        }
        if (b < arrmin[num - 1] + (l - arrmin[i]) * (base << num - i - 2)) l--;
        unsigned long long how = l - arrmin[i];
        arrmin[i] += how;
        unsigned long long sum = 0;
        for (int j = 0; j <= i; j++) sum += arrmin[j];
        for (int j = i + 1; j < num; j++) {
          arrmin[j] = sum + 1;
          sum += arrmin[j];
        }
      }
      for (int i = 0; i < num - 1; i++) cout << arrmin[i] << " ";
      cout << b << endl;
    } else
      cout << -1 << endl;
  }
  return 0;
}
