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;
const int INF = 0x3f3f3f3f;
const long long LINF = 0x3f3f3f3f3f3f3f3f;
const int MOD = 1e9 + 7;
template <typename T1, typename T2>
inline bool Min(T1 &a, const T2 &b) {
  if (b >= a) return false;
  a = b;
  return true;
}
template <typename T1, typename T2>
inline bool Max(T1 &a, const T2 &b) {
  if (b <= a) return false;
  a = b;
  return true;
}
template <typename T1, typename T2>
inline void Adm(T1 &a, const T2 &b) {
  a = (a + b) % MOD;
  if (a < 0) a += MOD;
}
long long x[100];
int main() {
  int T;
  cin >> T;
  while (T--) {
    long long a, b, m;
    scanf("%I64d%I64d%I64d", &a, &b, &m);
    if (a == b) {
      printf("1 %I64d\n", a);
      continue;
    }
    x[1] = a;
    for (int i = 2; i <= 50; ++i) {
      long long w = (1LL << i - 2);
      x[i] = 1;
      for (int j = 1; j < i; ++j) x[i] += x[j];
      if (x[i] > b) break;
      if (x[i] <= b && x[i] + w * (m - 1) >= b) {
        long long v = (b - x[i]) / w * w;
        for (int j = i; j > 1; --j) x[j] += v, v >>= 1;
        v = b - x[i];
        for (int j = i; j > 2; --j) {
          x[j] += v;
          if (v & 1) ++x[j - 1];
          v >>= 1;
        }
        printf("%d", i);
        for (int j = 1; j <= i; ++j) printf(" %I64d", x[j]);
        putchar('\n');
        goto brk;
      }
    }
    printf("-1\n");
  brk:;
  }
  return 0;
}
