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 MAXN = 1e6 + 10;
const int INF32 = 0x3f3f3f3f;
const long long INF64 = 0x3f3f3f3f3f3f3f3f;
const long long MOD = 1e9 + 7;
const double PI = acos(-1.0);
const double EPS = 1.0e-12;
long long R[55], Ans[55];
long long a, b, m;
int main() {
  int T;
  scanf("%d", &T);
  while (T--) {
    scanf("%lld%lld%lld", &a, &b, &m);
    if (a == b) {
      printf("1 %lld\n", a);
      continue;
    }
    if (b - a <= m) {
      printf("2 %lld %lld\n", a, b);
      continue;
    }
    int flag = 0;
    for (int k = 3; k <= 50; ++k) {
      long long c = b - a * (1ll << (k - 2));
      if (c < 0) break;
      long long l = 1ll << (k - 2);
      if (c < l) continue;
      c -= l;
      int f = 1;
      for (int i = 2; i <= k - 2; ++i) {
        long long tmp = c + (1ll << (k - i - 1));
        long long j = tmp / (1ll << (k - i - 1));
        R[i] = min(j, m);
        c = tmp - 1ll * R[i] * (1ll << (k - i - 1));
      }
      c += 2;
      if (c < 2 || c > 2ll * m) continue;
      R[k - 1] = c / 2;
      R[k] = c - R[k - 1];
      flag = k;
      break;
    }
    if (flag > 0) {
      printf("%d %lld", flag, a);
      long long sum = a;
      for (int i = 2; i <= flag; ++i) {
        long long res = sum + R[i];
        printf(" %lld", res);
        sum += res;
      }
      printf("\n");
      continue;
    }
    printf("-1\n");
  }
}
