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;
char _;
long long t;
long long a, b, m, s;
vector<long long> out;
bool check(long long sz) {
  out.clear();
  long long B = b;
  B -= (1LL << sz) * a;
  out.push_back(a);
  for (long long l = sz + 1 - 1; l >= 0; --l) {
    long long t = max(0LL, l - 1);
    long long tmp = min(m, B / (1LL << t));
    B -= (tmp) * (1LL << t);
    if (tmp == 0) {
      if (m == 1) {
        return 0;
      }
      if (l == 0)
        tmp = 1;
      else
        tmp = 2;
      long long pos, curr;
      for (pos = (long long)(out.size()) - 1, curr = 0; pos > 0 && curr == 0;
           --pos) {
        out[pos]--;
        curr = out[pos];
        if (out[pos] == 0) out[pos] = 2;
      }
      if (curr == 0) return 0;
    }
    out.push_back(tmp);
  }
  if (B != 0) return 0;
  return 1;
}
void run() {
  scanf("%lli%lli%lli", &a, &b, &m);
  if (a == b) {
    printf("1 %lli\n", a);
    return;
  }
  long long sz = 0;
  for (; (1LL << sz) * a <= b; ++sz) {
    if (check(sz)) {
      long long s = 0;
      printf("%i ", out.size());
      for (long long l = 0; l < out.size(); ++l) {
        cout << out[l] + s << ' ';
        s = s * 2 + out[l];
      }
      printf("\n");
      return;
    }
  }
  printf("-1\n");
}
int32_t main() {
  cin >> t;
  while (t--) {
    run();
  }
  return 0;
}
