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 double PI = acos(-1.0);
const long long MOD = 1e9 + 7;
const long long MAXN = 2e5 + 10;
const int INF = 0x3f3f3f3f;
const long long ll_INF = 9223372036854775807;
long long qm(long long a, long long b) {
  long long ret = 1;
  while (b) {
    if (b & 1) ret = ret * a % MOD;
    a = a * a % MOD;
    b >>= 1;
  }
  return ret;
}
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
long long lcm(long long a, long long b) { return (a * b) / gcd(a, b); }
long long a, b, m;
long long c[MAXN];
int main() {
  int T;
  scanf("%d", &T);
  while (T--) {
    scanf("%I64d%I64d%I64d", &a, &b, &m);
    if (a == b) {
      printf("1 %I64d\n", a);
      continue;
    }
    int flag = 1;
    long long l = a + 1, r = a + m;
    int n;
    if (l <= b && r >= b) {
      n = 2;
      printf("2 %I64d %I64d\n", a, b);
      continue;
    } else {
      for (int i = 3;; i++) {
        l = l * 2;
        r = r * 2;
        if (l <= b && r >= b) {
          n = i;
          break;
        }
        if (l > b) {
          flag = 0;
          break;
        }
      }
    }
    if (flag) {
      printf("%d", n);
      int num = 0;
      c[num] = b;
      int mid = 0;
      n -= 2;
      while (n--) {
        num++;
        if (c[num - 1] % 2 == 0) {
          c[num] = c[num - 1] / 2;
        } else {
          if (mid == 0) {
            c[num] = c[num - 1] / 2;
            mid = 1;
          } else {
            c[num] = (c[num - 1] + 1) / 2;
            mid = 0;
          }
        }
      }
      c[++num] = a;
      for (int i = num; i >= 0; i--) {
        printf(" %I64d", c[i]);
      }
    } else
      printf("-1");
    printf("\n");
  }
  return 0;
}
