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 power(long long int a, long long int b) {
  if (b == 0) return 1;
  long long int p = power(a, b / 2);
  if (b % 2 == 0)
    return p * p;
  else
    return a * p * p;
}
void shikhar7s(long long int cas) {
  long long int a, b, m, i, x, c, l, r;
  cin >> a >> b >> m;
  if (a == b) {
    cout << "1 " << a << "\n";
    return;
  }
  l = b / (a + m);
  r = b / (a + 1);
  for (i = 0; i < 55; i++) {
    c = power(2, i);
    if (c >= l && c <= r) {
      if (!(c == l && b % (a + m) != 0)) break;
    }
  }
  if (i == 55) {
    cout << -1 << "\n";
    return;
  }
  x = i + 1;
  b -= (c * a);
  vector<long long int> add(x, b / c);
  b %= c;
  vector<long long int> bi;
  while (b) {
    bi.push_back(b % 2);
    b /= 2;
  }
  long long int j = x - 2, bis = bi.size();
  for (i = 0; i < bis; i++) {
    add[j] += bi[i];
    j--;
  }
  cout << x + 1 << " " << a << " ";
  long long int s = a;
  l = a;
  for (i = 0; i < add.size(); i++) {
    s += add[i];
    cout << s << " ";
    s += l;
    l = s;
  }
  cout << "\n";
}
signed main() {
  std::ios_base::sync_with_stdio(false);
  long long int t = 1;
  cin >> t;
  long long int cas = 1;
  while (cas <= t) {
    shikhar7s(cas);
    cas++;
  }
  return 0;
}
