Problem: 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.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
using LL = long long;
const int MX = 55;
LL cur[MX], r[MX];
void print(int len) {
  cout << len << ' ';
  for (int i = 1; i <= len; i++) {
    LL ans = 0;
    for (int j = 1; j <= i; j++) ans += r[j] * cur[i - j];
    cout << ans << ' ';
  }
  cout << "\n";
}
void solve(LL a, LL b, LL m) {
  if (a == b) {
    cout << "1 " << a << "\n";
    return;
  }
  r[1] = a;
  for (int i = 2; i <= 50; i++) {
    LL rem = b - a * cur[i - 1];
    for (int j = 2; j <= i; j++) r[j] = 1, rem -= cur[i - j];
    if (rem < 0) break;
    for (int j = 2; j <= i; j++) {
      LL tmp = min(rem / cur[i - j], m - 1);
      r[j] += tmp;
      rem -= tmp * cur[i - j];
    }
    if (rem == 0) return print(i);
  }
  cout << "-1"
       << "\n";
}
int main() {
  cur[0] = 1;
  for (int i = 1; i <= 51; i++) cur[i] = 1LL << (i - 1);
  ios::sync_with_stdio(false);
  int T;
  cin >> T;
  while (T--) {
    LL a, b, m;
    cin >> a >> b >> m;
    solve(a, b, m);
  }
  return 0;
}