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;
void prnt(vector<long long int> a) {
  long long int n = a.size();
  for (long long int i = 0; i < n; i++) {
    long long int val = 0;
    for (long long int j = 0; j <= i; j++) val += a[j];
    cout << val << " ";
    a[i] = val;
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  long long int q;
  cin >> q;
  while (q--) {
    long long int a, b, m;
    cin >> a >> b >> m;
    if (a == b) {
      cout << 1 << " " << a << '\n';
    } else {
      for (long long int i = 0; i <= 48; i++) {
        long long int x = 1LL << i;
        long long int p = b / x, q = p;
        if (b % x) q++;
        if (p >= (a + 1) && q <= (a + m)) {
          long long int k = b - a * x - x;
          vector<long long int> ans;
          ans.push_back(a);
          for (long long int j = 1; j < i + 1; j++) {
            x /= 2;
            long long int y = k / x;
            if (y >= m) y = m - 1;
            ans.push_back(y + 1);
            k -= x * y;
          }
          ans.push_back(k + 1);
          cout << i + 2 << " ";
          prnt(ans);
          cout << '\n';
          goto ab;
        }
      }
      cout << -1 << '\n';
    ab:
      long long int dum;
    }
  }
}
