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;
template <typename T>
inline T gcd(T a, T b) {
  while (b != 0) swap(b, a %= b);
  return a;
}
template <typename T>
inline T lcm(T a, T b) {
  return ((a / gcd(a, b)) * b);
}
void solve() {
  unsigned long long int a, b, m;
  cin >> a >> b >> m;
  if (a == b) {
    cout << 1 << " " << a;
    return;
  }
  bool can = false;
  unsigned long long int k = 2, two = 1;
  for (; k <= 50; k++) {
    if (two * (a + 1) <= b && b <= two * (a + m)) {
      can = true;
      break;
    }
    two = two * 2;
  }
  if (!can) {
    cout << -1;
  } else {
    cout << k << " ";
    unsigned long long int n = (b / two) - a;
    unsigned long long int rem = b % two;
    vector<unsigned long long int> r(k + 1, 0);
    r[k] = n;
    for (unsigned long long int i = k - 1; i >= 2; i--) {
      r[i] = (n + rem % 2);
      rem = rem >> 1;
    }
    cout << a << " ";
    a = a + r[2];
    cout << a << " ";
    for (unsigned long long int i = 3; i <= k; i++) {
      a = 2 * a + r[i] - r[i - 1];
      cout << a << " ";
    }
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  unsigned long long int testnum = 1, totalnum = 1;
  cin >> totalnum;
  for (testnum = 1; testnum <= totalnum; ++testnum) {
    solve();
    cout << '\n';
  }
  return 0;
}
