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.
from sys import stdin, stdout

for q in range(int(input())):
    a, b, m = map(int, input().split())

    if a == b:
        print(1, a)
        continue

    k = 2

    while k <= 50 and not ((1 << (k-2)) * (a+1) <= b <= (1 << (k-2)) * (a+m)):
        k += 1
    
    if k > 50:
        print(-1)
        continue

    t = b - (1 << (k-2)) * (a+1)

    add = [1] * (k-1)
    add[0] = a

    for i in range(1, k-1):
        l = 1
        r = m
        while r - l > 1:
            x = (l + r) >> 1
            if t >= (1 << (k-2-i)) * (x-1):
                l = x
            else:
                r = x-1
        add[i] = r if t >= (1 << (k-2-i)) * (r-1) else l
        t -= (1 << (k-2-i)) * (add[i] - 1)

    ans = [a] * (k-1)
    for i in range(1, k-1):
        ans[i] = ans[i-1] * 2 - add[i-1] + add[i]
    print(k, ' '.join(str(i) for i in ans), b)
