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 q, a, b, m;
void solve() {
  cin >> a >> b >> m;
  if (a == b) {
    cout << 1 << ' ' << a << '\n';
    return;
  }
  if (a + m >= b) {
    cout << 2 << " " << a << " " << b << '\n';
    return;
  }
  for (long long sz = 3; sz <= 50; sz++) {
    long long bsa = (a << (sz - 2)), bb = b;
    if (bsa >= bb) {
      cout << -1 << '\n';
      return;
    }
    bool brk = 0;
    long long ad = 0;
    for (long long in = 2; in < sz; in++) {
      ad += ((long long)1 << (sz - in - 1));
      if (bsa + ad >= bb) {
        brk = 1;
        break;
      }
    }
    if (brk) continue;
    bsa += ad + 1;
    bb -= bsa;
    vector<long long> rmo(sz + 5, 0);
    for (long long in = 2; in < sz; in++) {
      long long cum = ((long long)bb >> (sz - in - 1));
      rmo[in] = min(m - 1, cum);
      bb -= (rmo[in] << (sz - in - 1));
    }
    rmo[sz] = min(bb, m - 1);
    bb -= rmo[sz];
    if (bb) continue;
    long long su = a;
    cout << sz << " " << a << " ";
    for (long long in = 2; in <= sz; in++) {
      cout << su + rmo[in] + 1 << " ";
      su += su + rmo[in] + 1;
    }
    cout << '\n';
    return;
  }
  cout << -1 << '\n';
  return;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> q;
  while (q--) {
    solve();
  }
  cout.flush();
  return 0;
}
