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>
const int maxn = 1e6 + 7;
const int mod = 1e9 + 7;
const double eps = 1e-8;
using namespace std;
long long x[maxn];
long long r[maxn];
int main() {
  int q;
  scanf("%d", &q);
  while (q--) {
    long long a, b, m;
    cin >> a >> b >> m;
    if (b == a) {
      cout << 1 << " " << a << endl;
      continue;
    }
    if (b == a + 1) {
      cout << 2 << " " << a << " " << b << endl;
      continue;
    }
    x[1] = a;
    x[2] = a + 1;
    int i;
    for (i = 3;; i++) {
      x[i] = 2ll * x[i - 1];
      if (x[i] > b) {
        i--;
        break;
      }
    }
    long long res = b - x[i];
    for (int j = 2; j < i; j++) {
      r[j] = 1;
      long long temp = min(m - 1, res / (1ll << (i - j - 1)));
      r[j] += temp;
      res -= 1ll * temp * (1ll << (i - j - 1));
    }
    r[i] = 1;
    long long temp = min(m - 1, res);
    r[i] += temp;
    res -= temp;
    if (res) {
      printf("%d\n", -1);
      continue;
    }
    x[1] = a;
    x[2] = a + r[2];
    for (int j = 3; j < i + 1; j++) {
      x[j] = r[j];
      for (int k = 1; k < j; k++) {
        x[j] += x[k];
      }
    }
    printf("%d ", i);
    for (int j = 1; j < i + 1; j++) {
      printf("%lld%c", x[j], j == i ? '\n' : ' ');
    }
  }
  return 0;
}