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 int em(long long int x, long long int Mod, long long int n) {
  if (n == 0)
    return 1ll;
  else if (n % 2 == 0)
    return em((x % Mod * x % Mod) % Mod, Mod, n / 2) % Mod;
  else
    return (x % Mod * em((x % Mod * x % Mod) % Mod, Mod, (n - 1) / 2)) % Mod;
}
int32_t main() {
  std::ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int t, i, j, k, l, a, b, m;
  cin >> t;
  while (t--) {
    cin >> a >> b >> m;
    if (a == b) {
      cout << 1 << " " << a << "\n";
      continue;
    }
    long long int lvl = 0, flag = 0;
    for (long long int i = (long long int)0; i < (long long int)50; i++) {
      long long int mini = (1ll << i) * (a + 1ll);
      long long int maxi = (1ll << i) * (a + m);
      if (b >= mini && b <= maxi) {
        lvl = i;
        break;
      }
      if (b < mini) {
        flag = 1;
        break;
      }
    }
    if (flag) {
      cout << "-1\n";
      continue;
    }
    vector<long long int> ans;
    ans.push_back(a);
    long long int sum = a;
    long long int now = 0;
    while (now != lvl) {
      long long int lo = 1, hi = m, res = -1;
      while (lo <= hi) {
        long long int mid = (lo + hi) / 2;
        long long int diff = lvl - now;
        long long int mini, maxi;
        mini = (mid + 1) * (1ll << (diff - 1)) + (1ll << diff) * sum;
        maxi = (mid + m) * (1ll << (diff - 1)) + (1ll << diff) * sum;
        if (b >= mini && b <= maxi) {
          res = max(res, mid);
          lo = mid + 1;
        } else if (mini < b)
          lo = mid + 1;
        else
          hi = mid - 1;
      }
      ans.push_back(sum + res);
      sum = 2ll * sum + res;
      now++;
    }
    ans.push_back(b);
    cout << lvl + 2 << " ";
    for (long long int y : ans) cout << y << " ";
    cout << "\n";
  }
}
