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 = input()
for _ in xrange(q):
    a, b, m = map(int, raw_input().strip().split())
    d = b - a
    if d >= 1 and d <= m:
        print 2, a, b
        continue

    if a == b:
        print 1, a
        continue

    found = 0
    nval = 3
    for n in xrange(3, 51):
        exp = n - 2
        lo = (a + 1) * (1 << exp)
        hi = (a + m) * (1 << exp)
        if lo <= b and b <= hi:
            found = 1
            nval = n
            break

    if not found:
        print -1
        continue
    
    n = nval

    pre = [1]
    cursum = 1
    while len(pre) < n:
        pre.append(cursum)
        cursum <<= 1

    r = [0] * (n - 1)

    pre.pop()

    or_b = b
    b = b - a * (1 << (n - 2))

    stexp = 1 << (n - 2) # starting exponent
    curb = 0 # current value of b filled
    for i in xrange(n - 2):
        stexp >>= 1
        stexp = max(stexp, 1)

        lo = ((b - curb - pre[-1 -i] * m) + stexp - 1) // stexp
        hi = (b - curb - pre[-1 -i]) // stexp

        ri = max(1, lo)
        ri = min(ri, hi)
        while 1:
            if ri <= 0:
                found = 0
                break

            bfill = ri * stexp

            left = b - (curb + bfill)
            if left >= pre[-1 -i] and left <= pre[-1 -i] * m:
                r[i] = ri
                curb += bfill
                break
            else:
                ri -= 1

        if found == 0:
            break

    if found == 0:
        print -1
        continue

    for i in xrange(n - 2):
        assert r[i] >= 1 and r[i] <= m
        
    print n, a,
    rsum = a
    x = a
    for i in xrange(n - 2):
        x = rsum + r[i]
        print x, 
        rsum += x
    assert or_b - rsum >= 1 and or_b - rsum <= m
    print or_b,
    print