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>
#pragma GCC optimize("O3")
using namespace std;
const int N = 2e5 + 5;
int q;
long long a, b, m;
vector<long long> v;
bool ok(int idx, long long x) {
  long long add = x, cur = 0;
  for (int j = idx + 1; j < v.size(); ++j) {
    if (v.back() + (add += cur) > b) return 0;
    cur = add;
  }
  return add + v.back() <= b;
}
int main() {
  scanf("%d", &q);
  while (q--) {
    scanf("%I64d%I64d%I64d", &a, &b, &m);
    v.clear();
    v.push_back(a);
    long long sum = a, sprv = 0;
    while (sum + sprv + 1 <= b) {
      long long tmp = sum;
      v.push_back(sum += 1 + sprv);
      sprv += tmp;
    }
    sum = v[0];
    for (int i = 1; i < v.size(); ++i) {
      long long st = 0, en = m - 1;
      while (st < en) {
        long long mid = (st + en + 1) / 2;
        if (ok(i, mid))
          st = mid;
        else
          en = mid - 1;
      }
      v[i] += st;
      sum += v[i];
      long long SUM = sum;
      for (int j = i + 1; j < v.size(); ++j) {
        v[j] = SUM + 1;
        SUM += v[j];
      }
    }
    if (v.back() != b)
      puts("-1");
    else {
      printf("%d", v.size());
      for (long long go : v) printf(" %I64d", go);
      puts("");
    }
  }
  return 0;
}
