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;
using ll = long long int;
void solve() {
  int q;
  cin >> q;
  while (q--) {
    ll a, b, m;
    cin >> a >> b >> m;
    if (a == b) {
      cout << 1 << '\n' << a << '\n';
      continue;
    }
    ll r[52];
    memset(r, -1, sizeof r);
    bool flag = 0;
    for (int i = 1; i < 50; ++i) {
      ll low = (a + 1) << (i - 1);
      if (low > b) break;
      ll high = (a + m) << (i - 1);
      if (b <= high) {
        flag = 1;
        ll pow = (1LL << (i - 1));
        ll rem = b - low;
        for (int pos = 1; pos <= i; ++pos) {
          pow /= 2;
          if (pow == 0) pow = 1;
          r[pos] = 1;
          if (pow) {
            r[pos] += min(m - 1, rem / pow);
            rem -= pow * (r[pos] - 1);
            rem = max(rem, 0LL);
          }
        }
        cout << i + 1 << '\n';
        r[0] = a;
        for (int pos = 0; pos <= i; ++pos) {
          for (int pos2 = 0; pos2 < pos; ++pos2) r[pos] += r[pos2];
          cout << r[pos] << ' ';
        }
        cout << '\n';
        break;
      }
    }
    if (!flag) cout << -1 << '\n';
  }
}
void InputSetup() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
}
int main(void) {
  auto start = chrono::high_resolution_clock::now();
  InputSetup();
  solve();
  auto finish = chrono::high_resolution_clock::now();
  cerr << "Time elapsed: "
       << (chrono::duration<long double>(finish - start)).count() << "s\n";
}
