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 MOD = 1e9 + 7;
vector<long long> f(long long a, long long b, long long m, int k) {
  vector<long long> ansBase(k);
  long long cs = a;
  ansBase[0] = a;
  for (int i = 1; i < k; i++) {
    ansBase[i] = cs + 1;
    cs += ansBase[i];
  }
  if (ansBase[k - 1] > b) {
    return vector<long long>();
  }
  if (ansBase[k - 1] == b) {
    return ansBase;
  }
  vector<long long> ans(k);
  ans[0] = a;
  cs = a;
  for (int i = 1; i < k; i++) {
    long long l = 1, r = m + 1;
    while (l < r) {
      long long mid = (l + r) / 2;
      long long val = cs + mid;
      long long cs2 = cs + val;
      for (int j = i + 1; j < k; j++) {
        val = cs2 + 1;
        cs2 += val;
        if (val > b) break;
      }
      if (val <= b)
        l = mid + 1;
      else
        r = mid;
    }
    ans[i] = cs + l - 1;
    cs += ans[i];
  }
  if (ans[k - 1] == b) return ans;
  return vector<long long>();
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int q;
  cin >> q;
  while (q--) {
    long long a, b, m;
    cin >> a >> b >> m;
    vector<long long> ans;
    for (int i = 1; i <= 50; i++) {
      auto a2 = f(a, b, m, i);
      if (a2.size()) {
        ans = a2;
        break;
      }
    }
    if (!ans.size()) {
      cout << "-1\n";
      continue;
    }
    cout << ans.size() << ' ';
    for (auto i : ans) {
      cout << i << ' ';
    }
    cout << '\n';
  }
  return 0;
}
