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 int maxn = 1e3 + 1;
long long n, i, res(0), j, a[maxn], q, t[maxn];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> q;
  n = 50;
  t[1] = 1;
  for (int i = 2; i <= n; ++i)
    for (int j = 1; j < i; ++j) t[i] += t[j];
  while (q--) {
    long long l, r, limit;
    cin >> l >> r >> limit;
    int i;
    if (l == r)
      cout << "1 " << l << "\n";
    else {
      vector<long long> test(n + 1, 0);
      test[1] = l;
      for (i = 2; i <= n; ++i) {
        for (int j = 1; j < i; ++j) test[i] += test[j];
        test[i]++;
        if (test[i] > r) break;
      }
      i--;
      long long need = r - test[i];
      int have = i;
      vector<long long> ans(i);
      vector<long long> mo(i, 0);
      for (int i = 0; i < ans.size(); ++i) ans[i] = test[i + 1];
      for (int i = 1; i < ans.size(); ++i) {
        long long remain = need / t[have - i];
        long long begins = 1, ends = limit - 1;
        while (begins <= ends) {
          long long mid = begins + ends >> 1;
          if (mid > remain)
            ends = mid - 1;
          else
            begins = mid + 1;
        }
        need -= ends * t[have - i];
        mo[i] = ends;
      }
      if (need > 0)
        cout << "-1\n";
      else {
        cout << ans.size() << " ";
        for (int i = 0; i < ans.size(); ++i) {
          for (int j = i; j < ans.size(); ++j) ans[j] += mo[i] * t[j - i + 1];
          cout << ans[i] << ' ';
        }
        cout << "\n";
      }
    }
  }
}
