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;
long long int mod = 1e9 + 7;
long long int max(long long int a, long long int b) {
  if (a > b)
    return a;
  else
    return b;
}
long long int min(long long int a, long long int b) {
  if (a < b)
    return a;
  else
    return b;
}
long long int binpow(long long a, long long b, long long m) {
  a %= m;
  long long res = 1;
  while (b > 0) {
    if (b & 1) res = res * a % m;
    a = a * a % m;
    b >>= 1;
  }
  return res;
}
long long binpow(long long a, long long b) {
  if (b == 0) return 1;
  long long res = binpow(a, b / 2);
  if (b % 2)
    return res * res * a;
  else
    return res * res;
}
void solve() {
  long long int a, b, m;
  cin >> a >> b >> m;
  vector<long long int> x(100);
  if (a == b) {
    cout << 1 << " " << a << endl;
    return;
  }
  long long int p = 1;
  int k = 2;
  while (b > p * (a + m)) {
    k++;
    p = p * 2;
  }
  if (b >= p * (a + 1)) {
    x[1] = a;
    x[k] = b;
    p = -1;
    for (int i = k - 1; i >= 2; i--) {
      if (x[i + 1] % 2 == 0) {
        x[i] = x[i + 1] / 2;
      } else {
        x[i] = (x[i + 1] + p) / 2;
        p = -p;
      }
    }
    cout << k << " ";
    for (int i = 1; i <= k; i++) {
      cout << x[i] << " ";
    }
    cout << endl;
    return;
  } else {
    cout << -1 << endl;
    return;
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int t;
  cin >> t;
  while (t--) {
    solve();
  }
}
