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;
long long q;
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cin >> q;
  while (q--) {
    long long a, b, m;
    cin >> a >> b >> m;
    vector<long long> X;
    bool found = false;
    for (long long k = 1; k <= 50; k++) {
      vector<long long> x;
      x.push_back(a);
      bool ok = 1;
      long long s = a;
      for (long long i = 0; i < k - 1; i++) {
        if (s + 1LL > b) {
          ok = 0;
          break;
        }
        x.push_back(s + 1LL);
        s += (s + 1LL);
      }
      if ((long long)x.size() != k or x.back() > b) ok = 0;
      if (!ok) continue;
      for (long long i = 1; i < k; i++) {
        long long nouv = 1LL;
        if (i < k - 1LL) nouv = (1LL << (k - 2LL - i));
        long long y = max(0LL, min(m - 1LL, (b - x.back()) / nouv));
        if (y > 0) {
          x[i] += y;
          for (long long j = i + 1LL; j < k; j++) {
            x[j] += y * (1LL << (j - i - 1LL));
          }
        }
      }
      if (x[0] != a or x.back() != b) ok = 0;
      s = a;
      for (long long i = 1; i < k; i++) {
        if (x[i] > s + m or x[i] < s + 1LL) {
          ok = 0;
          break;
        }
        s += x[i];
      }
      if (ok) {
        X.clear();
        found = true;
        for (long long i : x) {
          X.push_back(i);
        }
        break;
      }
    }
    if (found) {
      cout << (long long)X.size();
      for (long long i : X) cout << " " << i;
      cout << '\n';
    } else
      cout << -1 << '\n';
  }
}
