Problem: 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.

Solution (in PYTHON3):

Solution:
# from math import log2


def find_k(A, B, M):
    for k in range(2, 51):
        s = 2 ** (k-2)
        if (A+1) * s <= B and (A+M) * s >= B:
            return k
    return -1


def solve(A, B, M):
    if A == B:
        print('1', A)
        return
    k = find_k(A, B, M)
    if k == -1:
        print(-1)
        return
    print(k, end=' ')
    a = A
    print(A, end=' ')

    R = B - (2 ** (k-2)) * A
    for i in range(k-3, -1, -1):
        s = 2 ** i
        r = min(R // s - 1, M)
        print(a + r, end=' ')
        a += a + r
        R -= r * s
    assert 0 < R <= M
    print(B)


Q = int(input())
for case in range(Q):
    A, B, M = [int(x) for x in input().split()]
    solve(A, B, M)