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.
q = int(input())
for _ in range(q):
    a, b, m = [int(x) for x in input().split()]
    if a == b:
        print("1 ", a)
        continue
    found = 0
    for n in range(2, 51):
        lo = (2**(n-2))*(a + 1)
        hi = (2**(n-2))*(a + m)
        if lo <= b and b <= hi:
            b -= lo
            # print("b = ", b)
            print(n, a, "", end='')
            acc = a
            for i in range(2, n+1):
                p = 1 if i+1 >= n else (2**(n-i-1))
                left = 1
                right = m
                res = 1
                while left <= right:
                    mid = (left+right)//2
                    if p*mid-p <= b:
                        res = max(res, mid)
                        left = mid+1
                    else:
                        right = mid-1
                b+=p
                b-=p*res
                # print(i, p, res)
                print(acc+res, "", end='')
                acc += acc+res
            print("")
            found = 1
            break
    if not found:
        print(-1)
    