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 T;
const int maxn = 55;
long long ans[maxn];
long long Ans[maxn];
int main() {
  scanf("%d", &T);
  while (T--) {
    long long a, b, m;
    scanf("%lld%lld%lld", &a, &b, &m);
    if (a == b) {
      printf("1 %lld\n", a);
      continue;
    }
    memset(ans, 0, sizeof ans);
    bool f = 0;
    int cnt;
    for (int i = 1; i <= 50 && !f; i++) {
      for (int j = 0; j <= 52; j++) ans[j] = 0;
      long long c = b - (i - 2 < 0 ? (1ll) : (1ll << (i - 2))) * a;
      if (c <= 0) break;
      if (c > (i - 2 < 0 ? (1ll) : (1ll << (i - 2))) * m) continue;
      if (i > 1) c -= (i - 2 < 0 ? (1ll) : (1ll << (i - 2)));
      if (c < 0) break;
      for (int j = 1; j <= i - 1; j++) ans[j] = 1;
      for (int j = i - 3; j >= -1 && c; j--) {
        ans[i - 2 - j] += min(c / (j < 0 ? (1ll) : (1ll << (j))), m - 1);
        c = c - min(c / (j < 0 ? (1ll) : (1ll << (j))), m - 1) *
                    (j < 0 ? (1ll) : (1ll << (j)));
      }
      if (c == 0ll) {
        cnt = i;
        f = 1;
        break;
      }
    }
    if (!f)
      puts("-1");
    else {
      Ans[1] = a;
      for (int i = 1; i < cnt; i++) {
        long long sum = 0;
        for (int j = 1; j <= i; j++) sum += Ans[j];
        Ans[i + 1] = sum + ans[i];
      }
      printf("%d ", cnt);
      for (int i = 1; i <= cnt; i++) printf("%lld ", Ans[i]);
      puts("");
    }
  }
  return 0;
}
