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())
for i in range(q):
	a, b, m = list(map(int,input().split()))
	l = a + 1 
	r = a + m
	c = 0
	poziom = 0
	if a == b:
		print(1, a, end = "\n")
		continue
	while True:
		if l > b:
			print(-1)
			break
		if b <= r and b >= l:
			c = 1
			last = [l, r]
			break
		poziom += 1
		l *= 2
		r *= 2
	if c == 1:
		mat = [(a + 1) * 2 ** i for i in range(poziom + 1)]
		sumka = a
		for i in range(poziom + 1):
			if mat[-1] == b:
				break
			elif i == poziom:
				mat[-1] = b
				break
			else:
				pom = min(sumka + m - mat[i], (b - mat[-1]) // (2 ** (poziom - i - 1)))
				mat[i] += pom
				for j in range(i + 1, poziom + 1):
					mat[j] += pom * 2 ** (j - i -1)
				sumka += mat[i]
		print(len(mat) + 1, end = " ")
		print(a, end =  " ")
		for i in range(len(mat)):
			if i < len(mat) - 1:
				print(mat[i], end = " ")
			else:
				print(mat[i], end = "\n")
			
		