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;
const int sz = 60;
long long pow2[sz];
void solve(long long a, long long b, long long m) {
  int maxn = 55;
  for (int q = maxn; q > 0; q--) {
    long long cur = 0;
    vector<long long> r(q, 1);
    r[0] = a;
    vector<long long> coef(q, 1);
    for (int i = 0; i < q - 1; i++) {
      coef[i] = pow2[q - i - 2];
    }
    for (int i = 0; i < q; i++) {
      cur += coef[i] * r[i];
      if (coef[i] * r[i] < 0) {
        cur = -1;
        break;
      }
    }
    if (cur > b || cur < 0) continue;
    for (int i = 1; i < q; i++) {
      long long left = 1;
      long long right = m;
      long long mid;
      while (left <= right) {
        mid = (left + right) / 2;
        if (cur + (mid - 1) * coef[i] <= b && cur + (mid - 1) * coef[i] > 0) {
          r[i] = mid;
          left = mid + 1;
        } else {
          right = mid - 1;
        }
      }
      cur += (r[i] - 1) * coef[i];
    }
    if (cur == b) {
      cout << q << " ";
      long long sum = 0;
      for (int i = 0; i < q; i++) {
        long long pre = sum;
        sum += r[i];
        cout << sum << " ";
        sum += pre;
      }
      cout << "\n";
      return;
    }
  }
  cout << "-1\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  int q;
  cin >> q;
  pow2[0] = 1;
  for (int i = 1; i < sz; i++) {
    pow2[i] = pow2[i - 1] * 2;
  }
  while (q--) {
    long long a, b, m;
    cin >> a >> b >> m;
    solve(a, b, m);
  }
  return 0;
}
