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 q, a, b, m, k;
long long nb[100];
long long sum[100];
signed main() {
  cin >> q;
  while (q--) {
    memset(nb, 0, sizeof nb);
    cin >> a >> b >> m;
    if (a == b) {
      cout << "1 " << a << endl;
      continue;
    }
    long long pos = -1;
    for (register long long k = 2; k <= 50; k++) {
      long long pow = (1ll << (k - 2));
      if ((a + 1) * pow > b) break;
      if ((a + 1) * pow <= b && b <= (a + m) * pow) {
        pos = k;
        break;
      }
    }
    if (pos == -1) {
      cout << "-1" << endl;
      continue;
    }
    cout << pos << " " << a;
    long long remind = b - (a + 1) * (1ll << (pos - 2));
    m--;
    for (register long long i = pos, cnt = 2; remind; i--, cnt++) {
      long long val;
      if (i - 3 < 0)
        val = 1;
      else
        val = 1ll << (i - 3);
      long long num = remind / val;
      if (num <= m) {
        remind -= num * val;
        nb[cnt] = num;
      } else {
        remind -= m * val;
        nb[cnt] = m;
      }
    }
    sum[1] = a;
    for (register long long i = 2; i <= pos; i++) {
      long long tmp = 0;
      for (register long long j = 1; j < i; j++) tmp += sum[j];
      sum[i] = tmp + 1 + nb[i];
      cout << " " << sum[i];
    }
    cout << endl;
  }
  return 0;
}
