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 N = 50;
bool solve(long long a, long long b, long long m, int n) {
  long long num = (1LL << (n - 2));
  long long rem = (b % num);
  long long k = (b / num) - a;
  vector<long long> ans(n + 1), r(n + 1, k);
  for (int i = 2; i < n; ++i) {
    if ((rem & (1LL << (n - 1 - i)))) ++r[i];
  }
  ans[1] = a;
  long long sum = a;
  for (int i = 2; i <= n; ++i) {
    if (r[i] > m or r[i] < 1) return false;
    ans[i] = sum + r[i];
    sum += ans[i];
  }
  if (ans[n] != b) return false;
  cout << n << ' ';
  for (int i = 1; i <= n; ++i) cout << ans[i] << ' ';
  cout << '\n';
  return true;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  int q;
  cin >> q;
  while (q--) {
    long long a, b, m;
    cin >> a >> b >> m;
    if (a == b) {
      cout << 1 << ' ' << a << ' ' << '\n';
      continue;
    }
    bool found = false;
    for (int k = 2; k <= N; ++k) {
      if (solve(a, b, m, k)) {
        found = true;
        break;
      }
    }
    if (!found) cout << -1 << '\n';
  }
  return 0;
}
