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.
#include <bits/stdc++.h>
const int MAX_N = 100 + 10;
int q;
long long a, b, m, l[MAX_N], r[MAX_N];
long long qpow(long long a, long long t) {
  long long s = 1, tmp = a;
  for (; t; t >>= 1, tmp = tmp * tmp)
    if (t & 1) s = s * tmp;
  return s;
}
int main() {
  scanf("%d", &q);
  for (; q--;) {
    scanf("%lld %lld %lld", &a, &b, &m);
    if (a > b) {
      printf("-1\n");
      continue;
    }
    int o = 1;
    long long s = a;
    l[1] = a;
    for (; s + 1 <= b;) s += (l[++o] = s + 1);
    --m;
    if (!m && l[o] != b) {
      printf("-1\n");
      continue;
    }
    for (int i = 1; i <= o; ++i) r[i] = 0;
    long long cj = b - l[o];
    for (int i = 2; i <= o && cj; ++i) {
      long long ad = std::max(o - i - 1, 0), t;
      ad = qpow(2, ad);
      t = std::min(cj / ad, m);
      r[i] += t;
      cj -= t * ad;
    }
    if (cj > 0) {
      printf("-1\n");
      continue;
    }
    s = l[1];
    for (int i = 2; i <= o; ++i) s += (l[i] = s + r[i] + 1);
    printf("%d ", o);
    for (int i = 1; i <= o; ++i) printf("%lld ", l[i]);
    printf("\n");
  }
  return 0;
}
