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.
from itertools import count


#from hypothesis import given
#from hypothesis.strategies import integers
#
#
#ABM = integers(min_value=1, max_value=int(1e14))


def debug(*args, **kwargs):
    import sys
    #print(*args, *('{}: {}'.format(k, v) for k, v in kwargs.items()), sep='; ', file=sys.stderr)


def solve(a, b, m):
    if a == b:
        return [a]
    debug(a=a,b=b,m=m)
    for k in count(2):
        lo = 2**(k-2)*a + 2**(k-2)
        hi = 2**(k-2)*a + 2**(k-2)*m
        debug(k=k, lo=lo, hi=hi)
        if lo > b:
            break
        if b <= hi:
            ans = [a] + [2**(i-1)*a + 2**(i-1) for i in range(1, k)]
            debug("got one", start=ans)
            if m == 1:
                return ans
            for i in range(1, k - 1):
                deficit = b - ans[-1]
                quot = min(deficit // 2**(k - 2 - i), m - 1)
                ans[i] += quot
                for j in range(i + 1, k):
                    ans[j] += 2**(j - i - 1)*quot
            ans[-1] = b
            return ans
    return None


def check(a, b, m, ans):
    assert ans[-1] == b
    for i in range(1, len(ans)):
        assert 1 <= ans[i] - sum(ans[:i]) <= m


#@given(ABM, ABM, ABM)
def test_solve(a, b, m):
    if a > b:
        a, b = b, a
    ans = solve(a, b, m)
    if ans is None:
        return
    check(a, b, m, ans)


def main():
    q = int(input())
    for _ in range(q):
        a, b, m = map(int, input().split())


        ans = solve(a, b, m)
        debug(a=a, b=b, m=m, ans=ans)
        if ans:
            check(a, b, m, ans)
            print(len(ans), *ans)
        else:
            print(-1)


if __name__ == "__main__":
    main()
