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;
vector<long long> sol;
void solve(long long total, int left) {
  if (left <= 0) return;
  long long lo = total + 1;
  long long hi = total + m;
  while (lo <= hi) {
    long long next = (lo + hi) / 2;
    long long curA = next;
    long long sumA = total + next;
    long long curB = next;
    long long sumB = total + next;
    for (int i = 2; i <= left; i++) {
      curA = sumA + 1;
      sumA += curA;
      curB = sumB + m;
      sumB += curB;
      if (b < curA) break;
    }
    if (b >= curA && b <= curB) {
      sol.push_back(next);
      solve(total + next, left - 1);
      return;
    }
    if (b < curA)
      hi = next - 1;
    else
      lo = next + 1;
  }
}
int main() {
  scanf("%d", &q);
  while (q--) {
    scanf("%I64d%I64d%I64d", &a, &b, &m);
    if (a == b) {
      printf("1 %I64d\n", a);
      continue;
    }
    if (a > b) {
      printf("-1\n");
      continue;
    }
    long long curA = 0;
    long long sumA = a;
    long long curB = 0;
    long long sumB = a;
    for (int i = 2; i <= 50; i++) {
      curA = sumA + 1;
      sumA += curA;
      curB = sumB + m;
      sumB += curB;
      if (b >= curA && b <= curB) {
        sol.clear();
        sol.push_back(a);
        solve(a, i - 1);
        printf("%d ", sol.size());
        for (int j = 0; j < sol.size(); j++) {
          if (j) printf(" ");
          printf("%I64d", sol[j]);
        }
        printf("\n");
        break;
      }
      if (b < curA || i == 50) {
        printf("-1\n");
        break;
      }
    }
  }
}
