Problem: 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.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int q;
long long arr[55];
bool cannot;
bool check(long long st, long long ed, long long m, int n) {
  memset(arr, 0, sizeof arr);
  arr[1] = st;
  if (n == 1) return st == ed;
  long long dif = ed - st * (1ll << n - 2) - (1ll << n - 2);
  if (dif < 0) {
    cannot = true;
    return false;
  }
  long long sum = arr[1];
  for (int i = 2; i <= n; i++) {
    if (i == n) {
      if (dif < m) {
        arr[i] = sum + dif + 1;
        return true;
      }
      return false;
    }
    long long val = (1ll << n - i - 1);
    long long add = min(dif / val, m - 1);
    dif -= add * val;
    arr[i] = sum + add + 1;
    sum += arr[i];
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin >> q;
  while (q--) {
    long long a, b, m;
    cin >> a >> b >> m;
    bool flag = false;
    cannot = false;
    for (int len = 1; len <= 50; len++) {
      if (cannot) break;
      if (check(a, b, m, len)) {
        cout << len << ' ';
        for (int i = 1; i <= len; i++) cout << arr[i] << ' ';
        cout << endl;
        flag = true;
        break;
      }
    }
    if (!flag) cout << -1 << endl;
  }
  return 0;
}