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 main(void) {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long q;
  cin >> q;
  while (q--) {
    long long a, b, m;
    cin >> a >> b >> m;
    if (b == a) {
      cout << "1 " << a << "\n";
      continue;
    }
    long long r1 = a + 1, r2 = a + m;
    long long k = 1;
    while (r2 < b) {
      k++;
      r1 = r1 * 2;
      r2 = r2 * 2;
    }
    if (b < r1) {
      cout << "-1\n";
      continue;
    }
    long long k1 = k;
    long long p = b;
    long long pm;
    long long ans = k + 1;
    long long ansm[ans];
    long long rs[ans];
    long long tmp = b - a * pow(2, ans - 2) - pow(2, ans - 2);
    memset(rs, 0, sizeof rs);
    rs[0] = a;
    ansm[0] = a;
    for (__typeof(ans) i = (1) - ((1) > (ans)); i != (ans) - ((1) > (ans));
         i += 1 - 2 * ((1) > (ans))) {
      if (tmp == 0) break;
      long long pwt = ((ans - i - 2 >= 0) ? pow(2, ans - i - 2) : 1);
      rs[i] = min(tmp / pwt, m - 1);
      tmp -= (rs[i] * pwt);
    }
    if (tmp != 0) {
      cout << -1 << '\n';
      continue;
    }
    for (__typeof(ans) i = (1) - ((1) > (ans)); i != (ans) - ((1) > (ans));
         i += 1 - 2 * ((1) > (ans)))
      rs[i]++;
    for (__typeof(ans) i = (1) - ((1) > (ans)); i != (ans) - ((1) > (ans));
         i += 1 - 2 * ((1) > (ans))) {
      ansm[i] = rs[i];
      for (__typeof(i) j = (0) - ((0) > (i)); j != (i) - ((0) > (i));
           j += 1 - 2 * ((0) > (i)))
        ansm[i] += ansm[j];
    }
    cout << ans << ' ';
    for (__typeof(ans) i = (0) - ((0) > (ans)); i != (ans) - ((0) > (ans));
         i += 1 - 2 * ((0) > (ans)))
      cout << ansm[i] << ' ';
    cout << '\n';
  }
}
