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 MAX = 1e18;
const long long _mod = 1e9 + 7;
const long long N = 1e6 + 6;
int main() {
  ios::sync_with_stdio(false);
  int q;
  cin >> q;
  while (q--) {
    long long a, b, m;
    cin >> a >> b >> m;
    if (a == b) {
      cout << 1 << " " << a << endl;
      continue;
    }
    bool ok = true;
    long long a1 = 0, b1 = 0, a2 = a + 1, b2 = a + m;
    if (a2 <= b && b <= b2) {
      cout << 2 << " " << a << " " << b << endl;
      continue;
    }
    for (int i = 2; i <= 48; i++) {
      if (a2 >= MAX) break;
      a2 += a2;
      b2 += b2;
      if (a2 <= b && b <= b2) {
        ok = false;
        long long c = pow(2ll, i - 1) * a;
        c += 2;
        for (int j = 1; j < i - 1; j++) c += pow(2ll, i - j - 1);
        long long dd[60];
        for (int j = 1; j < 60; j++) dd[j] = 1;
        dd[0] = 0;
        if (c < b) {
          for (int j = 1; j < i - 1; j++) {
            long long tmp = pow(2ll, i - j - 1);
            long long dt = b - c;
            dd[j] += min(dt / tmp, m - 1);
            c += tmp * min(dt / tmp, m - 1);
            if (c == b) break;
          }
          if (c < b) {
            dd[i - 1] += min(m - 1, b - c);
            c += min(m - 1, b - c);
            if (b > c) dd[i] += b - c;
          }
        }
        long long s = a;
        cout << i + 1 << " " << a << " ";
        for (int j = 1; j <= i; j++) {
          cout << s + dd[j] << " ";
          s += s + dd[j];
        }
        cout << endl;
        break;
      }
    }
    if (ok) cout << "-1\n";
  }
  return 0;
}
