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>
const long long N = 50;
using namespace std;
long long power(long long b) {
  if (b == 0) return 1;
  if (b == 1) return 2;
  if (b % 2 == 1) return (power(b - 1) * 2);
  long long q = power(b / 2);
  return (q * q);
}
long long pwr[N];
void solve() {
  long long a, b, m, x = 0;
  cin >> a >> b >> m;
  if (a == b) {
    cout << "1 " << a << "\n";
    return;
  }
  for (long long i = 0; pwr[i] * (a + 1LL) <= b; ++i) {
    if (pwr[i] * (a + 1) <= b and b <= pwr[i] * (a + m)) {
      cout << i + 2 << " ";
      std::vector<long long> r(i + 1);
      b -= pwr[i] * (a + 1);
      for (long long j = 0; j < i; j++) {
        r[j] = b / pwr[i - j - 1];
        if (r[j] > m - 1) {
          r[j] = m - 1;
        }
        b -= pwr[i - j - 1] * r[j];
        r[j]++;
      }
      r[i] = b + 1;
      std::vector<long long> ans;
      long long sum = a;
      ans.push_back(a);
      for (long long j = 0; j < i + 1; j++) {
        ans.push_back(sum + r[j]);
        sum += ans[j + 1];
      }
      for (int j = 0; j < i + 2; ++j) {
        cout << ans[j] << " ";
      }
      cout << ""
           << "\n";
      return;
    }
  }
  cout << -1 << "\n";
  return;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long T = 1;
  cin >> T;
  for (int i = 0; i < N; ++i) {
    pwr[i] = power(i);
  }
  while (T--) {
    solve();
  }
  return 0;
}
