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;
const long long BIG = 1000000000000000000LL;
void solve();
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  long long tt;
  cin >> tt;
  for (long long i = (0); i < (tt); ++i) solve();
  return 0;
}
bool alerte = false;
bool tester(long long prem, long long der, long long ov, long long taille) {
  if (prem == der && taille == 1) {
    cout << "1 " << prem << "\n";
    return true;
  }
  if (taille == 1) return false;
  long long coeff = (1LL << (taille - 2));
  long long res = der - coeff * prem - coeff;
  if (res < 0) {
    alerte = true;
    return false;
  }
  vector<long long> raj(taille, 1);
  for (long long i = (1); i < (taille); ++i) {
    long long rem = (taille - 2) - i;
    if (rem < 0) rem = 0;
    long long ici = (1LL << rem);
    long long wo = res / ici;
    wo = min(wo, (ov - 1));
    raj[i] += wo;
    res -= wo * ici;
  }
  if (res != 0) return false;
  cout << taille << " " << prem << " ";
  long long som = prem;
  for (long long i = (1); i < (taille); ++i) {
    long long nv = som + raj[i];
    cout << nv << " ";
    som += nv;
  }
  cout << "\n";
  return true;
}
void solve() {
  long long a, b, m;
  cin >> a >> b >> m;
  alerte = false;
  for (long long k = (1); k < (51); ++k) {
    if (tester(a, b, m, k)) return;
    if (alerte) break;
  }
  cout << "-1\n";
}
