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 ans2[100], cnt[100];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  long long q, i, j, k, r;
  cin >> q;
  while (q--) {
    long long a, b, m;
    cin >> a >> b >> m;
    int fl = 0;
    if (a == b) {
      cout << 1 << " " << a;
      cout << "\n";
      continue;
    }
    for (i = 2; i <= 50; i++) {
      ans2[1] = a;
      long long s = a;
      int fl1 = 0;
      for (j = 2; j <= i; j++) {
        ans2[j] = s + 1;
        if (ans2[j] > b) {
          fl1 = 1;
          break;
        }
        s += ans2[j];
      }
      if (fl1) continue;
      long long diff = b - ans2[i];
      if (diff < 0) continue;
      for (j = 2; j <= i; j++) {
        cnt[j] = m - 1;
      }
      int T = 10;
      while (T--) {
        for (j = 51; j >= 0 && diff; j--) {
          long long s = (1LL << j);
          if (s == 1 && diff - 1 >= 0 && cnt[i] && i != 1) {
            long long z = min(diff / s, cnt[i]);
            ans2[i] += z;
            cnt[i] -= z;
            diff -= z;
            continue;
          }
          if (diff - s >= 0 && i - j - 1 >= 2 && cnt[i - j - 1]) {
            long long z = min(diff / s, cnt[i - j - 1]);
            diff -= s * z;
            cnt[i - j - 1] -= z;
            ans2[i - j - 1] += z;
            long long fg = 0;
            for (k = i - j; k <= i; k++) {
              ans2[k] += z * 1LL * (1LL << fg);
              fg++;
            }
          }
        }
      }
      if (diff) continue;
      cout << i << " ";
      for (j = 1; j <= i; j++) {
        cout << ans2[j] << " ";
      }
      cout << "\n";
      fl = 1;
      break;
    }
    if (!fl) cout << -1 << "\n";
  }
}
