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;
int q;
long long a, b, m;
bool can(long long startSum, long long low, long long high, long long target,
         long long add) {
  long long lowSum = startSum;
  long long highSum = startSum;
  while (high < target)
    lowSum += low, highSum += high, low = lowSum + 1LL, high = highSum + add;
  return low <= target && target <= high;
}
int main() {
  std::ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> q;
  for (int query = 1; query <= q; ++query) {
    cin >> a >> b >> m;
    if (!can(0, a, a, b, m)) {
      cout << -1 << '\n';
      continue;
    }
    long long sum = a;
    vector<long long> ans;
    ans.push_back(a);
    while (ans.back() != b) {
      long long low = sum + 1LL;
      long long high = sum + m + 1LL;
      while (low + 1LL != high) {
        long long mid = (low + high) / 2LL;
        if (can(sum, mid, high - 1, b, m))
          low = mid;
        else
          high = mid;
      }
      sum += low;
      ans.push_back(low);
    }
    cout << ans.size() << ' ';
    for (const auto& x : ans) cout << x << ' ';
    cout << '\n';
  }
  return 0;
}
