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 dp[55];
int main() {
  long long j = 1;
  for (int i = 2; i <= 52; i++) {
    dp[i] = j;
    j *= 2;
  }
  int q;
  scanf("%d", &q);
  while (q--) {
    long long a, b, m;
    scanf("%lld%lld%lld", &a, &b, &m);
    if (a == b) {
      printf("1 %lld\n", a);
      continue;
    }
    bool ok = 0;
    for (int i = 2; i <= 50; i++) {
      if (dp[i] * 1.0 * a >= 1e18) break;
      long long val = b - dp[i] * a;
      long long lo = dp[i], hi = m * dp[i];
      if (m * 1.0 * dp[i] >= 1e18) hi = 1e18;
      int k = i;
      if (lo <= val && val <= hi) {
        printf("%d %lld ", i, a);
        long long sum = a;
        for (int j = 2; j <= i - 1; j++) {
          long long low = 1, hii = m, res = 0;
          k--;
          lo = dp[k];
          while (low <= hii) {
            long long mid = (low + hii) / 2;
            long long lost = mid * dp[i - j + 1];
            if (mid * 1.0 * dp[i - j + 1] >= 1e18) lost = 1e18;
            if (lo <= val - lost) {
              res = mid, low = mid + 1;
            } else
              hii = mid - 1;
          }
          long long lost = res * dp[i - j + 1];
          val -= lost;
          printf("%lld ", sum + res);
          sum += sum;
          sum += res;
        }
        printf("%lld\n", b);
        ok = 1;
        break;
      }
    }
    if (!ok) printf("-1\n");
  }
}
