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.
import sys
range = xrange
input = raw_input

q = int(input())
for _ in range(q):
    a,b,m = [int(x) for x in input().split()]

    if a == b:
        print 1
        print a
        continue

    l = 1
    r = 50
    while l < r:
        mid = l + r + 1 >> 1
        
        LHS = b - (a << (mid - 2))
        RHS = 1 << (mid - 2)

        if LHS >= RHS:
            l = mid
        else:
            r = mid - 1

    if l == 1:
        print -1
        sys.exit()

    n = l

    LHS = b - (a << (n - 2))
    RHS = 1 << (n - 2)

    R = [1]*n
    R[0] = a
    for j in reversed(range(n - 2)):
        i = n - 2 - j
        
        power = 1 << j
        dx = min((LHS - RHS)//power, m - 1)
        R[i] += dx
        RHS += dx * power

    i = n - 1
    power = 1   
    dx = min((LHS - RHS)//power, m - 1)
    R[i] += dx
    RHS += dx * power

    X = [a]
    for i in range(1,n):
        X.append(sum(X) + R[i])
    
    if X[0] == a and X[-1] == b:
        print n, ' '.join(str(x) for x in X)
    else:
        print -1
