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;
int q;
long long a, b, m, mi[60], x[60], s[60], r[60];
int work(long long h, int n) {
  if (mi[n] > h) return 0;
  h -= mi[n];
  for (int i = 2; i <= n + 2; i++) r[i] = 1;
  r[n + 2] += min(m - 1, h);
  h -= min(m - 1, h);
  for (int i = n; i >= 1; i--) {
    long long k = min(m - 1, h / mi[i - 1]);
    r[n - i + 2] += k;
    h -= k * mi[i - 1];
  }
  if (h > 0) return 0;
  for (int i = 2; i <= n + 2; i++) {
    if (r[i] < 1 || r[i] > m) return 0;
  }
  return 1;
}
int main() {
  mi[0] = 1;
  for (int i = 1; i < 60; i++) mi[i] = mi[i - 1] << 1;
  cin >> q;
  while (q--) {
    cin >> a >> b >> m;
    if (a == b) {
      cout << "1 " << a << endl;
      continue;
    }
    if (b - a <= m) {
      cout << "2 " << a << ' ' << b << endl;
      continue;
    }
    int flag = 0;
    for (int i = 3; i <= 50; i++) {
      if (a * mi[i - 2] >= b) break;
      if (work(b - a * mi[i - 2], i - 2)) {
        flag = i;
        break;
      }
    }
    if (!flag)
      cout << "-1" << endl;
    else {
      x[1] = s[1] = a;
      cout << flag << ' ' << a;
      for (int i = 2; i <= flag; i++) {
        x[i] = s[i - 1] + r[i];
        s[i] = s[i - 1] + x[i];
        cout << ' ' << x[i];
      }
      cout << endl;
    }
  }
  return 0;
}
