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>
using namespace std;
const int maxn = 2e5 + 50;
int q;
long long a, b, m;
long long val[60];
long long tmp[60];
bool isSat(int k) {
  memcpy(tmp + 1, val + 1, k * sizeof(long long));
  for (int i = 2; i <= k; ++i) {
    long long x = m - 1;
    tmp[i] -= x;
    for (int j = i + 1; j <= k; ++j) {
      tmp[j] -= x;
      x += x;
    }
    if (tmp[k] <= b) return true;
  }
  return tmp[k] <= b ? true : false;
}
void Solve() {
  val[1] = a;
  long long sum = a;
  long long k = 1;
  while (k <= 50 && val[k] < b) {
    val[++k] = sum + m;
    sum += val[k];
  }
  if (k > 50 || !isSat(k)) {
    printf("-1\n");
    return;
  }
  printf("%lld %lld", k, a);
  for (int i = 2; i < k; ++i) {
    long long d = val[k] - b;
    long long x = min(m - 1, d / (1ll * 1 << (k - i - 1)));
    val[i] -= x;
    for (int j = i + 1; j <= k; ++j) val[j] -= x, x += x;
    printf(" %lld", val[i]);
  }
  if (k > 1)
    printf(" %lld\n", b);
  else
    printf("\n");
}
int main() {
  scanf("%d", &q);
  while (q--) {
    scanf("%lld%lld%lld", &a, &b, &m);
    Solve();
  }
  return 0;
}
