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
input = sys.stdin.readline

q=int(input())
Q=[list(map(int,input().split())) for i in range(q)]

for a,b,m in Q:

    MIN=[a]
    MAX=[a]
    SUMMIN=a
    SUMMAX=a
    flag=1

    for k in range(51):
        if MIN[-1]<=b<=MAX[-1]:
            break

        MIN.append(SUMMIN+1)
        MAX.append(SUMMAX+m)
        SUMMIN+=MIN[-1]
        SUMMAX+=MAX[-1]

        if MIN[-1]>10**14 or MIN[-1]>b:
            flag=0
            break

    if flag==0:
        print(-1)
        continue

    SA=b-MIN[-1]
    PLUS=[0]*k

    #print(SA,k)

    for j in range(k):
        x=SA//max(1,2**(k-j-2))
        if x>m-1:
            x=m-1
        PLUS[j]=x+1
        SA-=x*max(1,2**(k-j-2))

    #print(PLUS)

    ANS=[a]
    SUM=a
    for j in range(k):
        ANS.append(SUM+PLUS[j])
        SUM+=ANS[-1]

    print(len(ANS),*ANS)
    
        
    
        
        
