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 N = 200100;
const int M = 200100;
const int inf = 1e9;
const long double pi = 3.141592653589793238462643;
int n;
bool go(vector<long long>& v, long long a, long long b, long long m) {
  long long l = a;
  long long r = a;
  long long i = 0;
  while (r < b) {
    i++;
    l = (1ll << (i - 1)) * (a + 1);
    r = (1ll << (i - 1)) * (m + a);
  }
  if (b < l) return 0;
  vector<long long> ans;
  b -= (1ll << (i - 1)) * a;
  i--;
  while (i > 0) {
    long long f = (1ll << (i - 1));
    long long left = (1ll << (i - 1));
    long long right = (1ll << (i - 1)) * m;
    long long le = 1;
    long long ri = m + 1;
    long long mid = 1;
    while (1) {
      mid = (le + ri) / 2;
      if (b - f * mid < left)
        ri = mid;
      else if (b - f * mid > right)
        le = mid;
      else
        break;
    }
    b -= f * mid;
    ans.push_back(mid);
    i--;
  }
  ans.push_back(b);
  v.push_back(a);
  for (int i = 0; i < ((int)(ans).size()); i++) {
    long long cur = ans[i];
    for (int j = 0; j <= i; j++) cur += v[j];
    v.push_back(cur);
  }
  return 1;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 0; i < (n); i++) {
    long long a, b, m;
    cin >> a >> b >> m;
    vector<long long> v;
    if (a == b) {
      cout << 1 << " " << a << "\n";
      continue;
    }
    if (go(v, a, b, m)) {
      cout << ((int)(v).size()) << " ";
      for (int j = 0; j < (((int)(v).size())); j++) {
        cout << v[j] << " ";
      }
      cout << "\n";
    } else {
      cout << -1 << "\n";
    }
  }
  return 0;
}
