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;
int Q;
long long A, B, M;
bool f(int K) {
  if (K == 1) {
    if (A == B) {
      printf("%d %lld\n", 1, A);
      return true;
    }
    return false;
  }
  if (K == 2) {
    if (B - A <= M) {
      printf("%d %lld %lld\n", 2, A, B);
      return true;
    }
    return false;
  }
  long long a = A, b = (1LL << (K - 2));
  for (int i = 0; i < K - 2; i++) {
    a <<= 1;
    if (a + b > B) return false;
  }
  a = A + M;
  for (int i = 0; i < K - 2; i++) {
    a <<= 1;
    if (B <= a) {
      vector<long long> mul(51), r(51, 1), x(51);
      long long gap = B - (1LL << (K - 2)) * (A + 1);
      for (int j = 2; j <= K - 1; j++) mul[j] = (1LL << (K - j - 1));
      mul[K] = 1;
      for (int j = 2; j <= K; j++) {
        long long m = min(M - 1, gap / mul[j]);
        r[j] += m;
        gap -= m * mul[j];
      }
      x[1] = A;
      long long sum = A;
      for (int j = 2; j <= K; j++) {
        x[j] = sum + r[j];
        sum += x[j];
      }
      assert(x[K] == B);
      printf("%d ", K);
      for (int j = 1; j <= K; j++) printf("%lld ", x[j]);
      printf("\n");
      return true;
    }
  }
  return false;
}
int main() {
  scanf("%d", &Q);
  for (int q = 0; q < Q; q++) {
    scanf("%lld %lld %lld", &A, &B, &M);
    int chk = 0;
    for (int i = 1; i <= 50; i++)
      if (f(i)) {
        chk = 1;
        break;
      }
    if (chk == 0) printf("-1\n");
  }
  return 0;
}
