Problem: 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.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const long long maxn = 44;
long long u[55];
long long d[55];
long long R[55];
long long ans1[55];
long long ans2[55];
int main(void) {
  int T;
  scanf("%d", &T);
  while (T--) {
    long long a, b, m;
    scanf("%lld %lld %lld", &a, &b, &m);
    if (a == b) {
      printf("1 %lld\n", a);
      continue;
    }
    long long n = 0, flag = 1;
    u[1] = a;
    d[1] = a;
    long long flag2 = 0;
    long long i;
    for (i = 2; i <= 50; i++) {
      u[i] = flag * (a + m);
      d[i] = flag * (a + 1);
      if (b <= u[i] && b >= d[i]) {
        flag2 = i;
        break;
      }
      if (b < d[i]) {
        break;
      }
      flag *= 2;
    }
    if (flag2 == 0) {
      printf("-1\n");
      continue;
    } else {
      long long l = 1, r = m;
      while (l <= r) {
        long long mid = l + r >> 1;
        if (flag * (a + mid) < b) {
          l = mid + 1;
        } else if (flag * (a + mid) > b)
          r = mid - 1;
        else {
          r = mid;
          break;
        }
      }
      long long p = b - flag * (a + r);
      for (long long i = 1; i <= flag2; i++) R[i] = r;
      long long cnt = 1;
      while (p) {
        if (p & 1) {
          R[flag2 - cnt]++;
        }
        cnt++;
        p >>= 1;
      }
      ans1[1] = a;
      ans1[flag2] = b;
      ans2[1] = a;
      printf("%lld %lld", flag2, a);
      for (long long i = 2; i < flag2; i++) {
        ans1[i] = ans2[i - 1] + R[i];
        printf(" %lld", ans1[i]);
        ans2[i] = ans2[i - 1] + ans1[i];
      }
      printf(" %lld\n", b);
    }
  }
  return 0;
}