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 power(long long a, long long b, long long m = 998244353) {
  long long res = 1;
  while (b > 0) {
    if (b & 1) res = (res * a) % m;
    a = (a * a) % m;
    b >>= 1;
  }
  return res;
}
long long inverse(long long a, long long m = 998244353) {
  return power(a, m - 2, m);
}
void solve() {
  long long a, b, m;
  cin >> a >> b >> m;
  if (a == b) {
    cout << "1 " << a << endl;
    return;
  }
  int t, n = -1;
  long long pw = 1;
  for (t = 2; t <= 50; t++, pw *= 2) {
    long long mx = (a + m) * pw;
    long long mn = (a + 1) * pw;
    if (mn > b) break;
    if (mn <= b && mx >= b) {
      n = t;
      break;
    }
  }
  if (n == -1) {
    cout << -1 << endl;
    return;
  }
  b -= (a + 1) * pw;
  vector<long long> add;
  pw /= 2;
  while (pw > 0) {
    long long p = min(m - 1, b / pw);
    b -= p * pw;
    add.push_back(p);
    pw /= 2;
  }
  add.push_back(b);
  for (int i = 0; i < add.size(); i++) add[i]++;
  long long curr = a;
  cout << n << ' ';
  cout << curr << ' ';
  long long sum = curr;
  for (int i = 0; i < add.size(); i++) {
    curr = sum + add[i];
    cout << curr << ' ';
    sum += curr;
  }
  cout << endl;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int Q;
  cin >> Q;
  while (Q--) solve();
}
