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;
int const N = 3e6 + 10, oo = 1e9;
long long const OO = 2e18;
double const eps = 1e-8, PI = acos(-1);
int mod = oo + 7;
void solve(long long a, long long b, long long m) {
  if (a == b) {
    cout << 1 << ' ' << a << '\n';
    return;
  }
  if (b - a <= m) {
    cout << 2 << ' ' << a << ' ' << b << '\n';
    return;
  }
  for (int i = 3; i <= 50; i++) {
    vector<long long> tr;
    tr.push_back(a);
    for (int j = 1; j < i; j++) {
      tr.push_back(a * (1LL << (j - 1)) + (1LL << (j - 1)));
    }
    long long rem = b - tr.back();
    if (rem < 0) break;
    for (int j = 1; j < (int)tr.size(); j++) {
      long long ad = rem / (1LL << (i - j - 1 - (j != (int)tr.size() - 1)));
      ad = min(ad, m - 1);
      tr[j] += ad;
      for (int k = j + 1; k < (int)tr.size(); k++) {
        if (OO / (1LL << (k - j - 1)) < ad) break;
        tr[k] += ad * (1LL << (k - j - 1));
        if (k == (int)tr.size() - 1) rem -= ad * (1LL << (k - j - 1));
      }
    }
    if (tr.back() == b) {
      cout << i;
      for (int j = 1; j < (int)tr.size(); j++) {
        long long sum = 0;
        for (int k = 0; k < j; k++) {
          sum += tr[k];
        }
        if (tr[j] - sum > m) {
          cout << ' ' << tr[j] - sum << endl;
          assert(0);
        }
      }
      for (auto v : tr) cout << ' ' << v;
      cout << '\n';
      return;
    }
  }
  cout << -1 << '\n';
}
int main() {
  int q;
  cin >> q;
  while (q--) {
    long long a, b, m;
    cin >> a >> b >> m;
    solve(a, b, m);
  }
  return 0;
}
