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 long long int INF = 1e16;
int LIM = 1e5 + 10;
vector<vector<long long int> > adj(LIM);
long long int exp(long long int n, long long int e) {
  if (e == 0) return 1;
  long long int m = 1;
  if (n < 1e18 / n)
    m = n * n;
  else
    m = 1e18;
  long long int temp = exp(m, e / 2);
  if (e % 2 == 0)
    return temp;
  else {
    if (temp < (long long int)1e18 / n)
      return temp * n;
    else
      return 1e18;
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long int t;
  cin >> t;
  while (t--) {
    long long int ans = -1;
    long long int a, b, m;
    cin >> a >> b >> m;
    if (a == b) {
      cout << 1 << " " << a << endl;
      continue;
    }
    for (long long int i = 0; i <= 50; i++) {
      long long int temp = exp(2, i);
      if (temp > 1e15 / a) continue;
      temp = temp * a;
      assert(i >= 0 && i <= 50);
      long long int x = exp(2, i);
      if (1e15 / m <= x) continue;
      long long int y = x * m;
      if (b >= temp + x && b <= temp + y) {
        ans = i;
        break;
      }
    }
    if (ans == -1) {
      cout << -1 << endl;
      continue;
    }
    vector<long long int> vec;
    assert(ans >= 0 && ans <= 50);
    long long int c = b - exp(2, ans) * a;
    cout << ans + 2 << " " << a << " ";
    c = c * 2;
    assert(ans >= 0);
    for (long long int i = ans; i >= 0; i--) {
      long long int temp = c - c / 2 + exp(2, i) * a;
      vec.push_back(temp);
      c = c / 2;
    }
    reverse(vec.begin(), vec.end());
    for (long long int i = 0; i < vec.size(); i++) cout << vec[i] << " ";
    cout << endl;
  }
  return 0;
}
