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.
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#  code

if __name__ == "__main__":
    q = int(input())
    po = [1,1,2]
    l = 4
    for i in range(55):
        po.append(l)
        l *= 2
    while q > 0:
        q -= 1
        a,b,m = map(int,input().split())
        se = 0
        for ub in range(1,51):
            s = sum(po[:ub-1])
            if b < s + po[ub - 1] * a:
                print(-1)
                se = 2
                break
            if b <= s*m + po[ub - 1]*a :
                k = b - po[ub - 1]*a - po[ub - 1]
                arr = []
                for j in range(1,ub):
                    i = k//po[ub - 1 - j]
                    i = min(m - 1, i)
                    k -= i*po[ub - 1 - j]
                    arr.append(i + 1)

                print(ub,a,end=" ")
                s = a
                k = []
                for i in arr:
                    k.append(s + i)
                    s += (s + i)
                print(' '.join([str(i) for i in k]))
                se = 2
                break

        # print(arr)
        if se != 2:
            print(-1)
        
