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;
long long T, a, b, n, m, ans, num[200100], cf[60], f[60];
string str = "aeiou";
inline void Min(long long &u, long long v) {
  if (v < u) u = v;
}
inline void Max(long long &u, long long v) {
  if (v > u) u = v;
}
int main() {
  long long i, j, p, q, sum;
  f[0] = f[1] = 1;
  for (i = 2; i <= 50; i++) f[i] = f[i - 1] * 2;
  cin >> T;
  while (T--) {
    memset(cf, 0, sizeof(cf));
    scanf("%lld%lld%lld", &a, &b, &m);
    if (a == b) {
      printf("1 %lld\n", a);
      continue;
    }
    num[1] = sum = a;
    for (n = 1; num[n] <= b;) num[++n] = sum + 1, sum += num[n];
    n--, sum = b - num[n];
    for (i = 2; i <= n; i++) {
      long long t = min(sum / f[n - i], m - 1);
      cf[i] = t;
      sum -= t * f[n - i];
    }
    if (sum)
      puts("-1");
    else {
      printf("%lld %lld ", n, a);
      sum = a;
      for (i = 2; i <= n; i++) {
        num[i] = sum + cf[i] + 1;
        printf("%lld ", num[i]);
        sum += num[i];
      }
      puts("");
    }
  }
}
