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) {
  long long q, a, b, m;
  cin >> q;
  while (q--) {
    cin >> a >> b >> m;
    if (a == b) {
      printf("1 %lld\n", a);
      continue;
    }
    int flag = 0;
    for (long long i = 1, j = 1; a * i + i * 1 <= b; i += i, j++) {
      if (a * i + i * 1 <= b && b <= a * i + i * m) {
        stack<long long> q;
        long long t1 = a * i, t2 = b - a * i;
        long long t3 = t2;
        vector<long long> r;
        long long t = i;
        while (t2) {
          if (t == 1) {
            r.push_back(t2);
            break;
          }
          t = t / 2;
          if (t2 <= t * m) {
            t2 -= t;
            r.push_back(1);
          } else {
            long long x = (t2 - t * m) / t;
            if (x * t < t2 - t * m) {
              x++;
            }
            t2 -= t * x;
            r.push_back(x);
          }
        }
        while (j--) {
          q.push(t1 + t3);
          t1 -= t1 / 2;
          t3 -= r[j];
          long long c = 1;
          for (int k = j - 2; k >= 0; k--) {
            t3 -= c * r[k];
            c *= 2;
          }
        }
        printf("%d ", (int)q.size() + 1);
        printf("%lld", a);
        while (!q.empty()) {
          printf(" %lld", q.top());
          q.pop();
        }
        printf("\n");
        flag = 1;
        break;
      }
    }
    if (!flag) cout << "-1" << '\n';
  }
  return 0;
}
