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 pows[51];
long long sumpows[51];
void solve() {
  long long a, b, m;
  cin >> a >> b >> m;
  if (a == b) {
    cout << 1 << " " << a << "\n";
    return;
  }
  for (int terms = 2; terms < 51; terms++) {
    if (pows[terms] >= 1e18 / a) break;
    long long sum = pows[terms - 2] * a;
    int extra = terms - 2;
    if (sum >= 1e18 - sumpows[extra]) break;
    sum += pows[extra];
    if (b < sum) break;
    long long b2 = b - sum;
    vector<long long> ans(terms - 1, 1);
    for (long long i = terms - 2; i >= 1; i--) {
      long long po = pows[i - 1];
      long long req = b2 / po;
      long long coeff = min(m - 1, req);
      ans[terms - 2 - i] += coeff;
      b2 -= po * coeff;
    }
    if (b2 > m - 1) continue;
    ans.back() = b2 + 1;
    cout << terms << " ";
    long long cum = a;
    cout << a << " ";
    for (int i = 0; i < terms - 1; i++) {
      a = cum + ans[i];
      cout << a << " ";
      cum += a;
    }
    assert(a == b);
    cout << "\n";
    return;
  }
  cout << "-1\n";
  ;
}
int main() {
  pows[0] = 1;
  sumpows[0] = 1;
  for (int i = 1; i < 51; i++) {
    pows[i] = pows[i - 1] << 1;
    sumpows[i] = pows[i] + sumpows[i - 1];
  }
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  int t;
  cin >> t;
  while (t--) solve();
  return 0;
}
