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 pw[60];
long long a, b, m;
long long possible(long long in, long long fin, long long term) {
  if (term == 1) {
    if (in == fin)
      return 3;
    else
      return 2;
  }
  if (fin - in + 1 < term) {
    return 1;
  }
  long long baki = fin - pw[term - 2] * in;
  if ((baki > pw[term - 2] * m) || baki < pw[term - 2] || in > fin) {
    return 2;
  }
  return 3;
}
long long ajaira(long long mid, long long last, long long terms, long long sum,
                 long long koyta) {
  long long baki =
      last - pw[terms - koyta - 1] * sum - pw[terms - koyta - 2] * mid;
  if (baki > pw[terms - koyta - 2] * m) {
    return 1;
  } else if (baki < pw[terms - koyta - 2]) {
    return 2;
  } else {
    return 3;
  }
}
int main() {
  pw[0] = 1;
  for (long long i = 1; i < 59; i++) pw[i] = 2 * pw[i - 1];
  long long q;
  cin >> q;
  for (long long p = 1; p <= q; p++) {
    bool flag = false;
    cin >> a >> b >> m;
    if (a == b) {
      cout << 1 << ' ' << a << endl;
      continue;
    }
    for (long long k = 2; k <= 50; k++) {
      if (possible(a, b, k) == 3) {
        flag = true;
        long long ans[k];
        ans[0] = a;
        ans[k - 1] = b;
        long long sum = a;
        for (long long i = 1; i <= k - 2; i++) {
          long long low = 1;
          long long high = m;
          while (low <= high) {
            long long mid = (low + high) / 2;
            long long temp = ajaira(mid, b, k, sum, i);
            if (temp == 1) {
              low = mid + 1;
            } else if (temp == 2) {
              high = mid - 1;
            } else {
              if (ajaira(mid + 1, b, k, sum, i) != 3 || mid + 1 > m) {
                ans[i] = sum + mid;
                sum += ans[i];
                if (ans[i] > 100000000000000) {
                  flag = false;
                }
                break;
              } else {
                low = mid + 1;
              }
            }
          }
          if (ans[i] > 100000000000000) {
            flag = false;
          }
        }
        if (flag) {
          cout << k << ' ';
          for (long long i = 0; i < k; i++) {
            cout << ans[i] << ' ';
          }
          cout << endl;
          break;
        }
      }
    }
    if (!flag) {
      cout << -1 << endl;
      continue;
    }
  }
}
