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;
inline char getch() {
  static char buf[10000], *p1 = buf, *p2 = buf;
  return p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 10000, stdin), p1 == p2)
             ? EOF
             : *p1++;
}
char qwq;
inline long long read() {
  long long cz = 0, ioi = 1;
  qwq = getchar();
  while (qwq < '0' || qwq > '9')
    ioi = qwq == '-' ? ~ioi + 1 : 1, qwq = getchar();
  while (qwq >= '0' && qwq <= '9')
    cz = (cz << 3) + (cz << 1) + (qwq ^ 48), qwq = getchar();
  return cz * ioi;
}
long long n, pw[55], ans[55];
int main() {
  n = read();
  while (n--) {
    long long bj = 0, a = read(), b = read(), now = 1, c = read();
    if (a == b) {
      printf("1 %lld\n", a);
      continue;
    }
    for (register int len = 2; len <= 50; len++) {
      long long qaq = b - a * now - now;
      if (qaq < 0) break;
      long long nw = now;
      for (register int i = 2; i < len; i++) {
        nw /= 2;
        long long cs = min(c - 1, qaq / nw);
        qaq -= nw * cs;
        ans[i] = cs;
      }
      if (qaq >= c) {
        now *= 2;
        continue;
      }
      ans[len] = qaq;
      bj = 1;
      long long sum = a;
      cout << len << " " << a << " ";
      for (register int i = 2; i <= len; i++) {
        ans[i]++;
        ans[i] += sum;
        sum += ans[i];
        cout << ans[i] << " ";
      }
      cout << endl;
      break;
    }
    if (!bj) puts("-1");
  }
}
