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 main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long int i, j;
  long long int n, k;
  int test;
  cin >> test;
  while (test--) {
    long long int a, b, m;
    cin >> a >> b >> m;
    vector<pair<long long int, long long int> > v;
    v.push_back({a, a});
    k = 1;
    long long int fs = a, ss = a;
    while (k <= 49) {
      if ((fs + 1) > b) break;
      v.push_back({fs + 1, ss + m});
      fs = 2 * fs + 1;
      ss = 2 * ss + m;
      k++;
    }
    if (a == b) {
      cout << 1 << " " << a << "\n";
      continue;
    }
    if ((a + m) >= b) {
      cout << 2 << " " << a << " " << b << "\n";
      continue;
    }
    int check = 0;
    for (i = 0; i < v.size(); i++) {
      if (v[i].second >= b && b >= v[i].first) {
        check = 1;
        break;
      }
    }
    if (check == 0)
      cout << "-1\n";
    else {
      long long int fm, fl = v.size();
      fm = (b - (pow(2, fl - 2) * a)) * 1.00 / (pow(2, fl - 2));
      long long int diff = b - pow(2, fl - 2) * (a)-pow(2, fl - 2) * (fm);
      vector<long long int> ans, bit, ps(fl + 10, 0);
      while (diff != 0) {
        bit.push_back(diff % 2);
        diff = diff / 2;
      }
      long long int sb = bit.size();
      for (i = 1; i <= (fl - 1 - sb); i++) bit.push_back(0);
      reverse(bit.begin(), bit.end());
      for (i = 1; i <= 50; i++) bit.push_back(0);
      ps[0] = a + bit[0];
      ans.push_back(a + bit[0]);
      k = 1;
      long long int fs = a + bit[0];
      for (i = 1; i <= fl - 1; i++) {
        ans.push_back(fs + fm + bit[i]);
        ps[i] = ps[i - 1] + ans[i];
        fs = ans[i] + ps[i - 1];
        k++;
      }
      cout << ans.size() << " ";
      for (i = 0; i < ans.size(); i++) {
        cout << ans[i] << " ";
      }
      cout << "\n";
    }
  }
  return 0;
}
