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 mul[55], pre[55];
long long r[55];
long long dd[55];
long long anc[55];
int main() {
  mul[1] = mul[2] = 1;
  pre[1] = 1;
  pre[2] = 2;
  for (int i = 3; i <= 50; i++) {
    mul[i] = mul[i - 1] * 2;
    pre[i] = pre[i - 1] + mul[i];
  }
  int q;
  scanf("%d", &q);
  while (q--) {
    long long a, b, m;
    scanf("%lld%lld%lld", &a, &b, &m);
    if (a == b) {
      printf("1 %lld\n", a);
      continue;
    } else if (b - a <= m) {
      printf("2 %lld %lld\n", a, b);
      continue;
    }
    bool f = 0;
    long long tmp = b - a;
    for (int i = 3; i <= 50; i++) {
      r[i] = r[i - 1] = 1;
      for (int j = i - 2; j >= 2; j--) {
        r[j] = r[j + 1] * 2;
      }
      long long x = (mul[i] - 1);
      long long y = 0;
      for (int j = 2; j <= i; j++) y += r[j];
      long long sum1 = x * a + y;
      long long sum2 = x * a + y * m;
      if (tmp >= sum1 && tmp <= sum2) {
        tmp -= x * a;
        f = 1;
        for (int j = 2; j <= i; j++) {
          dd[j] = 1;
          tmp -= r[j] * dd[j];
        }
        for (int j = 2; j <= i; j++) {
          if (tmp == 0) break;
          if (tmp >= r[j] && dd[j] < m) {
            long long cnt = min(tmp / r[j], m - dd[j]);
            dd[j] += cnt;
            tmp -= cnt * r[j];
          }
        }
        anc[1] = a;
        anc[2] = a + dd[2];
        for (int j = 3; j <= i; j++) {
          anc[j] = 2LL * anc[j - 1] - dd[j - 1] + dd[j];
          if (anc[j] > 100000000000000) f = 0;
        }
        if (!f) continue;
        printf("%d", i);
        for (int j = 1; j <= i; j++) printf(" %lld", anc[j]);
        puts("");
        break;
      }
    }
    if (!f) puts("-1");
  }
  return 0;
}
