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.
def solve():
    a, b, m = (int(x) for x in input().split())
    if a == b:
        print(1, a)
        return
    curr = a
    curr_const = 1
    steps_nr = 0
    while b - curr > curr_const * m:
        curr *= 2
        curr_const *= 2
        steps_nr += 1
    if b - curr < curr_const:
        print(-1)
        return


    left = b - curr - curr_const
    ans = [a]
    sum = a
    curr_const //= 2
    while curr_const > 0:
        curr_try = left // curr_const
        curr_try = min(m - 1, curr_try)
        left -= curr_try * curr_const
        ans.append(sum + curr_try + 1)
        sum += ans[-1]
        curr_const //= 2

    curr_const = 1
    curr_try = left // curr_const
    curr_try = min(m - 1, curr_try)
    left -= curr_try * curr_const
    ans.append(sum + curr_try + 1)
    sum += ans[-1]
    curr_const //= 2

    print(len(ans), *ans)



if __name__ == '__main__':
    q = int(input())
    for _ in range(q):
        solve()