Problem: 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.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
void solve(long long a, long long b, long long m) {
  long long k = 0;
  for (; ((a + 1) << (k + 1)) <= b; k++) {
  }
  long long rem = b - ((a + 1) << k);
  vector<long long> v;
  long long t = a;
  v.push_back(a);
  for (long long r = k - 1; r >= 0; r--) {
    long long bef = t;
    t = t * 2 + 1;
    long long p = rem / (1LL << r);
    if (p > m - 1) p = m - 1;
    t += p;
    rem -= (1LL << r) * p;
    v.push_back(t - bef);
  }
  if (b - t <= m) {
    v.push_back(b);
    cout << v.size();
    for (int i = 0; i < v.size(); i++) cout << " " << v[i];
    cout << endl;
  } else
    cout << -1 << endl;
}
int main() {
  long long q;
  cin >> q;
  for (int i = 0; i < q; i++) {
    long long a, b, m;
    scanf("%lld%lld%lld", &a, &b, &m);
    if (a == b) {
      cout << 1 << " " << a << endl;
    } else
      solve(a, b, m);
  }
  return 0;
}