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;
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  long long t;
  cin >> t;
  long long it = t;
  while (t--) {
    long long a, b, m;
    cin >> a >> b >> m;
    long long len = 0, mn = 0;
    if (a == b) {
      cout << "1 " << a << "\n";
      continue;
    }
    for (long long i = 2; i <= 50; ++i) {
      long long pref = a, cur = 0;
      bool work = true;
      for (long long j = 2; j <= i; ++j) {
        cur = pref + 1;
        if (cur > b) {
          work = false;
          break;
        }
        pref += cur;
      }
      long long lb = cur;
      if (!work) continue;
      cur = 0;
      pref = a;
      for (long long j = 2; j <= i; ++j) {
        cur = pref + m;
        if (cur >= b) break;
        pref += cur;
      }
      work &= cur >= b;
      if (work) {
        len = i;
        mn = lb;
      }
    }
    if (len == 0) {
      cout << "-1\n";
      continue;
    }
    long long left = b - mn;
    long long rr[51];
    fill(rr, rr + 51, 1);
    for (long long i = 1; i < len - 1; ++i) {
      long long pw = 1;
      long long pow = len - 2 - i;
      for (long long j = 1; j <= pow; ++j) {
        pw *= 2;
      }
      if (left < pw) continue;
      long long l = 1, r = m - 1;
      while (l < r) {
        long long md = l + (r - l + 1) / 2;
        if (md > 1e15 / pw) {
          r = md - 1;
          continue;
        }
        if (md * pw > left) {
          r = md - 1;
        } else
          l = md;
      }
      rr[i] += l;
      left -= pw * l;
    }
    rr[len - 1] += left;
    vector<long long> ans;
    long long cur = 0, pref = a;
    ans.push_back(a);
    for (long long i = 1; i < len; ++i) {
      cur = pref + rr[i];
      ans.push_back(cur);
      pref += cur;
    }
    cout << len << " ";
    for (long long i : ans) {
      cout << i << " ";
    }
    cout << "\n";
  }
}
