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())

def prepare_sequence(a, b, curr_two_pow, m):
  sequence = [a]
  curr_pow = 1
  diff = b-curr_two_pow*a
  while curr_two_pow > 1:
    m_candidate = 1
    diff -= m_candidate

    while diff > curr_two_pow//2: 
      m_candidate*=2
      diff = diff-(m_candidate-m_candidate//2)

    if diff < curr_two_pow//2:
      m_candidate //=2
      diff = diff + (m_candidate)

    curr_candidate = m_candidate//(curr_two_pow//2)
    if curr_candidate > m:
      diff = diff + ((curr_candidate-m)*(curr_two_pow//2))
      curr_candidate = m
  
    sequence.append(sum(sequence)+curr_candidate)

    curr_pow *= 2
    curr_two_pow //= 2

  sequence.append(b)

  return sequence
  

def satisfies(a, b, m):
  if a == b:
    return [a]
  curr_two_pow = 1 
  while curr_two_pow*a < b:
    diff = b-curr_two_pow*a
    if diff >= curr_two_pow and diff <= curr_two_pow*m:
      return prepare_sequence(a, b, curr_two_pow, m)
    curr_two_pow *= 2
  return None

for i in range(q):
  a, b, m = map(int, input().split())
  res = satisfies(a,b,m)
  if res:
    print("%s %s" % (len(res), " ".join(map(str, res))))
  else:
    print(-1)

