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 solve() {
  long long a, b, m;
  cin >> a >> b >> m;
  for (long long i = 0; i < 50; i++) {
    long long k = ((long long)1 << ((max((long long)0, i - 1))));
    if (!i) {
      if (a == b) {
        cout << 1 << " " << a << '\n';
        return;
      }
    } else {
      if (b >= k * (a + 1) && b <= k * (a + m)) {
        vector<long long> r(i + 1);
        r[0] = 0;
        b -= k * (a + 1);
        for (long long j = 1; j <= i; j++) {
          long long t = (long long)1 << max((long long)0, i - j - 1);
          long long p = min(b / t, m - 1);
          b -= t * p;
          r[j] = p + 1;
        }
        cout << i + 1 << " ";
        long long cur = 0;
        r[0] = a;
        for (long long j = 0; j <= i; j++) {
          cout << cur + r[j] << " ";
          cur += cur + r[j];
        }
        cout << '\n';
        return;
      }
    }
    if (b < k * a) {
      break;
    }
  }
  cout << -1 << '\n';
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int tc;
  tc = 1;
  cin >> tc;
  for (int i = 1; i <= tc; i++) {
    solve();
  }
  return 0;
}
