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;
const long long inf = 2LL * 1e18;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int q;
  cin >> q;
  while (q--) {
    long long a, b, m;
    cin >> a >> b >> m;
    if (a == b) {
      cout << 1 << " " << a << endl;
      continue;
    }
    int len;
    for (len = 2; len <= 50; ++len) {
      if (b / (a + 1) >= (1LL << len - 2) &&
          (b + a + m - 1) / (a + m) <= (1LL << len - 2))
        break;
    }
    if (len > 50) {
      cout << "-1" << endl;
      continue;
    }
    b -= (1LL << len - 2) * (a + 1);
    --len;
    vector<long long> r(len, 0);
    for (int i = 0; i < len; ++i) {
      if (len - i >= 2) {
        long long L = 0, R = m - 1;
        while (L < R) {
          long long mid = (L + R + 1) >> 1;
          if ((1LL << len - i - 2) > b / mid) {
            R = mid - 1;
          } else {
            L = mid;
          }
        }
        b -= (1LL << len - i - 2) * L;
        r[i] = L;
      } else {
        r[i] = b;
      }
    }
    vector<long long> ans;
    ans.push_back(a);
    ans.push_back(a + 1 + r[0]);
    for (int i = 1; i < len; ++i)
      ans.push_back(2 * ans[ans.size() - 1] + r[i] - r[i - 1]);
    cout << ans.size();
    for (long long num : ans) cout << " " << num;
    cout << endl;
  }
  return 0;
}
