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 MAXN = 100;
long long coef[MAXN];
int main() {
  int Q;
  scanf("%d", &Q);
  while (Q--) {
    memset(coef, 0, sizeof(coef));
    bool flag = true;
    long long a, b, m;
    scanf("%lld%lld%lld", &a, &b, &m);
    if (a == b) {
      printf("1 %lld\n", a);
      continue;
    }
    long long tt = a + m;
    int kk = 0;
    while (tt < b) {
      tt <<= 1;
      kk++;
    }
    long long rr = b - (a << kk);
    if (rr < 0)
      flag = false;
    else {
      bool mk = false;
      long long q = rr >> kk;
      long long r = rr - (q << kk);
      for (int i = 0; i <= kk; i++) {
        coef[i] = q;
        if (i == 0 && r > 0) {
          coef[0]++;
          r--;
        } else {
          if (r & 1) coef[i]++;
          r >>= 1;
        }
      }
    }
    for (int i = 0; i <= kk; i++) {
      if (coef[i] == 0) {
        flag = false;
        break;
      }
    }
    if (!flag)
      puts("-1");
    else {
      printf("%d %lld", kk + 2, a);
      for (int i = 0; i <= kk; i++) {
        printf(" %lld", a + coef[kk - i]);
        a = (a << 1) + coef[kk - i];
      }
      putchar('\n');
    }
  }
  return 0;
}
