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 = 3e5 + 5;
const int mod = 1e9 + 7;
long long qpow(long long a, long long b) {
  b = max(b, 0ll);
  long long ans = 1;
  a %= mod;
  assert(b >= 0);
  for (; b; b >>= 1) {
    if (b & 1) ans = ans * a;
    a = a * a;
  }
  return ans;
}
long long gcd(long long a, long long b) { return b > 0 ? gcd(b, a % b) : a; }
long long a, b, m;
int check(long long x) {
  long long l = qpow(2, x - 1) * (a + 1);
  long long r = qpow(2, x - 1) * (a + m);
  if (l <= b && b <= r)
    return 0;
  else if (r > 0 && b > r)
    return -1;
  return 1;
}
int main() {
  long long q, k;
  cin >> q;
  while (q--) {
    cin >> a >> b >> m;
    long long l = 1, r = 1, mid;
    if (a == b) {
      cout << 1 << " " << b << endl;
      continue;
    }
    while (qpow(2, r - 1) * a < b) r++;
    while (l < r) {
      mid = (l + r) / 2;
      k = check(mid);
      if (k == 0)
        r = mid;
      else if (k > 0)
        r = mid;
      else
        l = mid + 1;
    }
    if (check(l) != 0 || l + 1 > 50) {
      cout << -1 << endl;
    } else {
      long long L = l + 1, temp = qpow(2, l - 1);
      l = 0, r = b / temp;
      while (l <= r) {
        mid = (l + r) / 2;
        k = b - temp * (a + mid);
        if (k < 0)
          r = mid - 1;
        else if (k <= temp) {
          l = mid;
          r = b - temp * (a + mid);
          break;
        } else if (k > temp) {
          l = mid + 1;
        }
      }
      bitset<30> x = r;
      cout << L << " " << a << " ";
      long long aa, bb, cc;
      for (int i = 1; i < L; i++) {
        aa = (r >> (L - 1 - i));
        bb = ((r >> (L - 2 - i)) & 1);
        cc = qpow(2, i - 1) * (a + l);
        cout << qpow(2, i - 1) * (a + l) + aa + bb << " ";
      }
      cout << endl;
    }
  }
  return 0;
}
