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 N = 300010;
const long long mod = 998244353;
long long dp[55][2];
vector<long long> v;
void output() {
  cout << v.size();
  for (int i = (int(v.size()) - 1); i >= (0); i--) {
    cout << " " << v[i];
  }
  cout << endl;
}
void solve(long long a, long long b, long long m) {
  v.clear();
  dp[1][0] = dp[1][1] = a;
  int n = 0;
  for (int i = (1); i <= (50); i++) {
    n = i;
    if (i > 1) {
      dp[i][0] = dp[i - 1][0] + 1;
      dp[i][1] = dp[i - 1][1] + m;
    }
    if (b < dp[i][0]) {
      cout << -1 << endl;
      return;
    }
    if (b >= dp[i][0] && b <= dp[i][1]) break;
    dp[i][0] += dp[i - 1][0];
    dp[i][1] += dp[i - 1][1];
  }
  long long now, val;
  val = b - dp[n - 1][1];
  if (val >= 1 && val <= m) {
    now = b + dp[n - 1][1];
  } else
    now = b + b - 1;
  while (n > 1) {
    val = now - dp[n - 1][1] * 2;
    if (val >= 1 && val <= m) {
      v.push_back(now - dp[n - 1][1]);
      now = dp[n - 1][1];
    } else {
      v.push_back(now - (now - 1) / 2);
      now = (now - 1) / 2;
    }
    n--;
  }
  v.push_back(a);
  output();
}
int main() {
  int n;
  cin >> n;
  long long a, b, m;
  for (int i = (1); i <= (n); i++) {
    cin >> a >> b >> m;
    solve(a, b, m);
  }
  return 0;
}
