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;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  long long q;
  cin >> q;
  for (long long i = 0; i < q; i++) {
    long long a, b, m;
    cin >> a >> b >> m;
    if (m == 1) {
      vector<long long> x((size_t)50, a);
      for (long long j = 1; j < 50; j++) {
        x[j] = accumulate(x.begin(), x.begin() + j, 0LL) + 1;
      }
      auto it = find(x.begin(), x.end(), b);
      if (it == x.end()) {
        cout << -1 << endl;
      } else {
        cout << it - x.begin() + 1 << " ";
        for (auto jt = x.begin(); jt <= it; jt++) {
          cout << *jt << " ";
        }
        cout << endl;
      }
    } else {
      bool found = false;
      for (long long j = 1; j <= 50; j++) {
        if (j == 1) {
          if (a == b) {
            cout << 1 << " ";
            cout << a << " ";
            cout << endl;
            found = true;
            break;
          }
          continue;
        }
        if (j == 2) {
          if (b - a <= m) {
            cout << 2 << " ";
            cout << a << " ";
            cout << b << " ";
            cout << endl;
            found = true;
            break;
          }
          continue;
        }
        long long d = b - a * (1ULL << (j - 2ULL)) - (1ULL << (j - 2ULL));
        if (d >= 0) {
          vector<long long> v;
          for (long long k = 0; k < j - 1; k++) {
            long long h = (k == j - 2 ? 1 : (1ULL << (j - 3ULL - k)));
            long long g = min(d / h, m - 1);
            v.push_back(g);
            d -= g * h;
          }
          if (d == 0) {
            vector<long long> x((size_t)j, a);
            bool filled = true;
            for (long long k = 1; k < j; k++) {
              x[k] = accumulate(x.begin(), x.begin() + k, 0LL) + v[k - 1] + 1;
              if (x[k] > (long long)1e14) {
                filled = false;
                break;
              }
            }
            if (filled) {
              cout << j << " ";
              for (long long k : x) {
                cout << k << " ";
              }
              cout << endl;
              found = true;
              break;
            }
          }
        }
      }
      if (!found) {
        cout << -1 << endl;
      }
    }
  }
}
