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.
#!/Users/genaro/venvs/venv3/bin/python

import math

n = int(input())

def solve(a, b, m):
    if 1 <= (b-a) <=m:
        return [2, a, b]
    if a == b:
        return [1, a]
    x = math.log2(b / (a + m))
    # print(x)
    y = math.log2(b / (a + 1))
    # print(y)
    if math.ceil(x) > math.floor(y):
        return [-1]
    mi = math.floor(y)
    w = mi
    k = b - 2 ** mi * (a + 1)
    # print(k, w)
    r = [w + 2, a]
    s = a
    while True:
        inc = 1
        if k > 0:
            if w > 0:
                _inc = int(min (m -1, k // 2 ** (w-1) ) )
                # print(_inc)
                k -= _inc * 2 ** (w-1)
                inc += _inc
            else:
                inc += k
        r.append(s + inc)
        s += r[-1]
        # print(w, inc, k, r)
        if r[-1] == b:
            break
        w -= 1

        if w == -2:
            break
    return r

for _ in range(n):
    a, b, m = list(map(int, input().split()))
    r = solve(a,b,m)
    s = ' '.join(map(str, r))
    print(s)
