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;
using ll = long long;
using pii = pair<int, int>;
constexpr ll N = 1e5 + 1;
constexpr ll INF = (1LL << 32) + 7;
vector<ll> ans, diff;
void f(ll a, ll b, ll m) {
  ans.clear();
  diff.clear();
  if (b == a) {
    ans.push_back(a);
    return;
  }
  if (b - a <= m) {
    ans.push_back(a);
    ans.push_back(b);
    return;
  }
  ll el = 1, sum = 0, temp = 0, lel;
  while (el <= b / a) {
    if (b - el * a <= el * m && b - el * a >= el) {
      lel = el;
      sum = b - el * a;
      el >>= 1;
      while (el) {
        temp = min((sum - el) / el, m);
        sum -= el * temp;
        diff.push_back(temp);
        el >>= 1;
      }
      temp = min(sum, m);
      diff.push_back(temp);
      ans.push_back(a);
      temp = a + diff[0];
      for (ll i = 1; i < (ll)diff.size(); i++) {
        ans.push_back(temp);
        temp += temp + diff[i] - diff[i - 1];
      }
      ans.push_back(b);
      return;
    }
    el <<= 1;
  }
  ans.push_back(-1);
}
void Solve() {
  ll q, a, b, m;
  cin >> q;
  while (q--) {
    cin >> a >> b >> m;
    f(a, b, m);
    if (ans[0] != -1) cout << ans.size() << " ";
    for (auto i : ans) cout << i << " ";
    cout << "\n";
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  Solve();
  return 0;
}
