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 ft, lt;
long long lim;
long long pow2[61];
long long r[100];
long long a[100], sum[100];
int main() {
  cin >> q;
  pow2[0] = 1;
  for (int i = 1; i <= 60; i++) {
    pow2[i] = pow2[i - 1] << 1;
  }
  while (q--) {
    cin >> ft >> lt >> lim;
    if (ft == lt) {
      printf("1 %lld\n", lt);
      continue;
    }
    bool fa = 0;
    for (int n = 2; n <= 50; n++) {
      r[1] = ft;
      long long x = lt;
      x -= (1ll << (n - 2)) * r[1];
      for (int i = 2; i < n; i++) {
        r[i] = 1;
        x -= (1ll << (n - i - 1));
      }
      r[n] = 1;
      x--;
      if (x < 0) {
        break;
      }
      if (x < -1) {
        n--;
        r[n] *= 4;
      }
      for (int i = 2; i <= n; i++) {
        if (i == n) break;
        r[i] += x / (1ll << (n - i - 1));
        if (r[i] >= lim) {
          r[i] = lim;
        }
        x -= pow2[n - i - 1] * (r[i] - 1);
      }
      r[n] += x;
      bool is_ok = 1;
      for (int i = 2; i <= n; i++) {
        if (r[i] < 1 || r[i] > lim) {
          is_ok = 0;
          break;
        }
      }
      if (is_ok) {
        printf("%d ", n);
        for (int i = 0; i <= n; i++) {
          if (i == 0) {
            continue;
          }
          a[i] = sum[i - 1] + r[i];
          sum[i] = sum[i - 1] + a[i];
          printf("%lld ", a[i]);
        }
        cout << "\n";
        fa = 1;
        break;
      }
    }
    if (!fa) {
      puts("-1");
    }
  }
  return 0;
}
