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 power(long long x, long long y, long long md = 1000000007) {
  long long res = 1;
  x %= md;
  while (y > 0) {
    if (y & 1) res = (res * x) % md;
    x = (x * x) % md;
    y = y >> 1;
  }
  return res % md;
}
signed main() {
  long long t;
  cin >> t;
  long long p[59];
  p[0] = 1;
  for (long long i = 1; i < 51; i++) p[i] = p[i - 1] * 2;
  while (t--) {
    long long a, b, m;
    cin >> a >> b >> m;
    if (a == b) {
      cout << 1 << " " << a << '\n';
      continue;
    }
    long long second = a, term = a;
    std::vector<long long> v;
    map<long long, long long> v1;
    v.push_back(term);
    long long first = 1;
    for (long long i = 1; i <= 50; i++) {
      term = second + 1;
      v.push_back(term);
      second += term;
      if (term > b) break;
      if (b - term <= ((m - 1) * p[i - 1])) {
        long long diff = b - term;
        for (long long j = 1; j < i; j++) {
          if (diff == 0) break;
          if (diff >= p[i - j - 1]) {
            long long x = diff / p[i - j - 1];
            x = min(x, m - 1);
            diff -= x * p[i - j - 1];
            v1[j] = x;
          }
        }
        if (diff) v1[i] = diff;
        long long sum = a;
        for (long long j = 1; j < v.size(); j++) {
          v[j] = sum + v1[j] + 1;
          sum += v[j];
        }
        cout << v.size() << " ";
        for (long long j = 0; j < v.size(); j++) cout << v[j] << " ";
        cout << '\n';
        first = 0;
        break;
      }
    }
    if (first) cout << -1 << '\n';
  }
}
