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;
long long pow(int k) {
  if (k == -1) {
    return 1;
  } else if (k >= 0) {
    return 1LL << k;
  } else {
    return 0;
  }
}
bool solve(long long A, long long B, long long M, int k) {
  if (k >= 3 && ((B >> (k - 2)) > A + M || (B >> (k - 2)) < A + 1)) {
    return false;
  }
  long long lb = (A + 1) * pow(k - 2);
  long long ub = (A + M) * pow(k - 2);
  if (lb <= B && ub >= B) {
    long long over = B - lb;
    vector<long long> add(k - 1, 1);
    for (int j = 0; j < k - 1; ++j) {
      long long d = min(over / pow(k - 3 - j), M - 1);
      add[j] += d;
      over -= d * pow(k - 3 - j);
    }
    printf("%d %lld", k, A);
    for (int j = 0; j < k - 1; ++j) {
      printf(" %lld", A + add[j]);
      A += A + add[j];
    }
    printf("\n");
    return true;
  } else {
    return false;
  }
}
int main() {
  int Q;
  scanf("%d", &Q);
  for (int q = 0; q < Q; ++q) {
    long long A, B, M;
    scanf("%lld%lld%lld", &A, &B, &M);
    if (A == B) {
      printf("1 %lld\n", A);
      continue;
    }
    bool solved = false;
    for (int k = 2; k <= 50; ++k) {
      if (solve(A, B, M, k)) {
        solved = true;
        break;
      }
    }
    if (!solved) {
      printf("-1\n");
    }
  }
  return 0;
}
