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 int r[70];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int TESTS = 1;
  cin >> TESTS;
  long long int a[100];
  a[0] = 1;
  for (int i = 1; i <= 55; i++) {
    a[i] = a[i - 1] * 2;
  }
  while (TESTS--) {
    long long int x, y, m;
    cin >> x >> y >> m;
    if (x == y) {
      cout << 1 << " " << x << '\n';
      continue;
    }
    long long int n = 0;
    for (int i = 1; i <= 55; i++) r[i] = 0;
    for (int i = 2; i <= 50; i++) {
      long long int p = a[i - 2] * x + a[i - 2];
      long long int q = a[i - 2] * x + a[i - 2] * m;
      if (y < p) break;
      if (y >= p && y <= q) {
        n = i;
        break;
      }
    }
    if (n == 0) {
      cout << -1 << '\n';
      continue;
    }
    long long int gap = y - a[n - 2] * x;
    long long int ox;
    for (int i = 2; i <= n; i++) r[i] = 1;
    gap -= a[n - 2];
    for (int i = n - 3; i >= 0; i--) {
      ox = gap / a[i];
      ox = min(ox, m - 1);
      gap -= ox * a[i];
      r[n - i - 1] += ox;
    }
    r[n] += gap;
    long long int ans[105];
    ans[1] = x;
    ans[2] = ans[1] + r[2];
    for (int i = 3; i <= n; i++) {
      ans[i] = 2 * ans[i - 1] + r[i] - r[i - 1];
    }
    cout << n << " ";
    for (int i = 1; i <= n; i++) {
      cout << ans[i] << " ";
    }
    cout << '\n';
  }
  return 0;
}
