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 read() {
  long long x = 0, f = 1;
  char ch = getchar();
  while (ch - '0' < 0 || ch - '0' > 9) {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch - '0' >= 0 && ch - '0' <= 9) {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
int Q;
long long st, en, m;
long long a[55];
long long cha[55];
int main() {
  Q = read();
  while (Q--) {
    st = read();
    en = read();
    m = read();
    if (st == en) {
      printf("1 %lld\n", st);
      continue;
    }
    a[1] = st;
    long long sum = st;
    int len = 1;
    while (true) {
      if (sum + 1 > en) break;
      len++;
      a[len] = sum + 1;
      sum += a[len];
    }
    for (int i = 1; i < len; i++) cha[i] = 1;
    long long tmp = en - a[len];
    for (int i = len - 3; i >= 0; i--) {
      long long k = min(tmp / (1ll << i), m - 1);
      cha[len - i - 2] += k;
      tmp -= k * (1ll << i);
      if (tmp == 0) break;
    }
    if (tmp <= m - 1)
      cha[len - 1] += tmp;
    else {
      puts("-1");
      continue;
    }
    printf("%d ", len);
    sum = st;
    printf("%lld ", st);
    for (int i = 1; i < len; i++) {
      printf("%lld ", sum + cha[i]);
      sum += sum + cha[i];
    }
    puts("");
  }
  return 0;
}
