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 power(long long a, long long n) {
  if (n == 1) return a;
  if (n == 0) return 1;
  if (n % 2) return a * power(a * a, n / 2);
  return power(a * a, n / 2);
}
const long long md = 998244353;
const long long inf = (long long)1e18;
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  long long q;
  cin >> q;
  while (q--) {
    long long a, b, m, fo = 0;
    cin >> a >> b >> m;
    if (a == b) {
      cout << "1 " << a << "\n";
      continue;
    }
    vector<long long> ans;
    for (long long i = 2; i < 51; ++i) {
      if (i == 2) {
        if (b - a >= 1 && b - a <= m) {
          ans.push_back(a);
          ans.push_back(b - a);
          fo = 1;
          break;
        }
        continue;
      }
      long long tot;
      if (b / a >= (1LL << (i - 2)))
        tot = b - (1LL << (i - 2)) * a;
      else
        break;
      vector<long long> tmp;
      tmp.push_back(a);
      for (long long j = 3; j < i + 1; ++j)
        tot -= (1LL << (i - j)), tmp.push_back(1);
      tot--;
      tmp.push_back(1);
      if (tot < 0) continue;
      for (long long j = 1; j < (long long)tmp.size() - 1; ++j) {
        tot += (1LL << (i - j - 2));
        long long x = min(tot / (1LL << (i - j - 2)), m);
        tmp[j] = x;
        tot -= (x * (1LL << (i - j - 2)));
      }
      if (tot + 1 <= m) {
        tmp.back() = tot + 1;
        ans = tmp;
        fo = 1;
        break;
      }
    }
    if (fo) {
      cout << (long long)ans.size() << " ";
      for (long long i = 0; i < (long long)ans.size(); ++i) {
        if (i > 0) ans[i] += ans[i - 1];
        cout << ans[i] << " ";
        if (i > 0) ans[i] += ans[i - 1];
      }
      cout << "\n";
    } else
      cout << "-1\n";
  }
  return 0;
}
