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 int INF = 1001001001;
const long long LINF = 1e16;
const double pi = 3.1415926535897932;
const string endstr = "\n";
template <typename T>
T gcd(T a, T b) {
  return (a == 0) ? b : gcd(b % a, a);
}
template <typename T>
T lcm(T a, T b) {
  return a / gcd(a, b) * b;
}
bool p_comp_fs(const pair<long long, long long> p1,
               const pair<long long, long long> p2) {
  return p1.first < p2.first;
};
bool p_comp_fg(const pair<long long, long long> p1,
               const pair<long long, long long> p2) {
  return p1.first > p2.first;
};
bool p_comp_ss(const pair<long long, long long> p1,
               const pair<long long, long long> p2) {
  return p1.second < p2.second;
};
bool p_comp_sg(const pair<long long, long long> p1,
               const pair<long long, long long> p2) {
  return p1.second > p2.second;
};
template <typename T>
vector<T> uniquen(vector<T> vec) {
  vec.erase(unique(vec.begin(), vec.end()), vec.end());
  return vec;
}
int main() {
  long long Q;
  cin >> Q;
  for (long long q = 0; q < Q; q++) {
    long long a, b, m;
    cin >> a >> b >> m;
    bool done = false;
    if (b == a) {
      cout << 1 << " " << a << endl;
      continue;
    }
    for (long long i = (1); i < 51; i++) {
      if (done) break;
      long long coffa = i <= 2 ? 1ll : 1ll << (i - 2);
      if (b <= coffa * a) {
        cout << -1 << endl;
        break;
      }
      long long rem = b - coffa * a;
      long long j = i - 1;
      vector<long long> ms;
      long long remj = rem - (j == 0 ? 1ll : 1ll << (j - 1));
      if (remj < 0) {
        cout << -1 << endl;
        break;
      }
      for (long long k = j - 2; k >= 0; k--) {
        long long dm = 1ll << k;
        long long qt = min(remj / dm, m - 1);
        ms.push_back(qt);
        remj -= dm * qt;
      }
      long long qt = min(remj, m - 1);
      ms.push_back(qt);
      remj -= qt;
      if (remj == 0) {
        cout << ms.size() + 1 << " " << a;
        long long sum = a;
        for (long long k = 0; k < ms.size(); k++) {
          long long rec = sum + ms[k] + 1;
          sum += rec;
          cout << " " << rec;
        }
        cout << endl;
        done = true;
        break;
      }
    }
  }
  return 0;
}
