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 int get(long long int x) {
  long long int y = 1, ans = 0;
  while (y < x) {
    y = y * 2;
    if (y <= x)
      ans++;
    else
      break;
  }
  return ans;
}
int main() {
  int t = 1;
  cin >> t;
  while (t--) {
    long long int n, m, k, a = 0, b = 0, ans = 1, x = 0, y = 0;
    cin >> a >> b >> m;
    if (a == b) {
      cout << "1 " << a << "\n";
      continue;
    }
    n = get(b / (a + 1));
    long long int arr[51] = {};
    long long int rem = b - (1ll << n) * (a + 1);
    long long int cur = a + 1, sum = a;
    vector<long long int> sol;
    sol.push_back(n + 2);
    sol.push_back(a);
    for (int i = 0; i < n; i++) {
      x = (1ll << (n - i - 1));
      if (rem >= x) {
        k = min(m - 1, rem / x);
        rem -= k * x;
        cur += k;
      }
      sol.push_back(cur);
      sum += cur;
      cur = sum + 1;
    }
    if (rem >= 0 && rem < m) {
      sol.push_back(cur + rem);
      for (auto i : sol) cout << i << " ";
    } else {
      cout << "-1";
    }
    cout << "\n";
  }
  return 0;
}
