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;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int q;
  cin >> q;
  while (q--) {
    ll a, b, m;
    cin >> a >> b >> m;
    if (a == b) {
      cout << 1 << " " << a << "\n";
      continue;
    }
    bool f = false;
    for (int k = 1; k <= 49 && (1ll << (k - 1)) <= b / a; k++) {
      ll t = (1ll << (k - 1)) * a;
      ll b_tmp = b - t;
      vector<ll> rs(k + 1);
      rs[0] = a;
      for (int cur = 1; cur <= k; cur++) {
        ll pw = (cur == k) ? 1 : (1ll << (k - cur - 1));
        rs[cur] = 1;
        b_tmp -= pw * rs[cur];
      }
      if (b_tmp < 0) {
        continue;
      }
      for (int cur = 1; cur <= k; cur++) {
        ll pw = (cur == k) ? 1 : (1ll << (k - cur - 1));
        ll now = min(m - 1, b_tmp / pw);
        b_tmp -= pw * now;
        rs[cur] += now;
      }
      if (b_tmp == 0) {
        f = true;
        cout << k + 1 << " ";
        vector<ll> ans;
        for (int i = 0; i <= k; i++) {
          ll cur = rs[i];
          for (int j = i - 1; j >= 0; j--) {
            cur += ans[j];
          }
          ans.push_back(cur);
          cout << cur << " ";
        }
        cout << "\n";
        break;
      }
    }
    if (!f) cout << -1 << "\n";
  }
  return 0;
}
