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 = 0x3f3f3f3f;
const int mod = 10007;
const int maxn = 1e4 + 5;
const double PI = acos(-1.0);
const double E = 2.718281828459;
const double eps = 1e-7;
int main() {
  int T;
  cin >> T;
  while (T--) {
    long long a, b, m;
    cin >> a >> b >> m;
    if (a == b) {
      cout << 1 << " " << a << endl;
      continue;
    }
    long long xn = b;
    long long n = -1;
    for (int i = 2; i < 60; i++) {
      if (b < (1ll << (i - 2)) * a) break;
      long long sub = b - (1ll << (i - 2)) * a;
      long long Min = 1ll << (i - 2);
      long long Max = (1ll << (i - 2)) * m;
      if (sub >= Min && sub <= Max) {
        n = i;
        break;
      }
    }
    if (n == -1) {
      cout << n << endl;
      continue;
    }
    long long res[maxn];
    b -= (1ll << (n - 2)) * a;
    b -= 1ll << (n - 2);
    for (int i = 2; i <= n - 1; i++) {
      long long p2 = 1ll << (n - i - 1);
      long long g = b / p2;
      if (g > m - 1) g = m - 1;
      b -= g * p2;
      res[i] = g + 1;
    }
    res[n] = b + 1;
    long long ans[maxn];
    ans[1] = a;
    long long sum[maxn];
    sum[1] = a;
    cout << n << " " << a << " ";
    for (int i = 2; i <= n; i++) {
      ans[i] = sum[i - 1] + res[i];
      sum[i] = sum[i - 1] + ans[i];
      cout << ans[i] << " ";
    }
    printf("\n");
  }
  return 0;
}
