Problem: 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.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int N = 105;
const long long MAX = 1e14;
long long q, a, b, m;
long long r[N];
long long qp(long long A, long long B) {
  long long ans = 1;
  while (B) {
    if (B & 1) ans = ans * A;
    A = A * A;
    B >>= 1;
  }
  return ans;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> q;
  while (q--) {
    memset(r, 0, sizeof(r));
    cin >> a >> b >> m;
    if (a == b) {
      cout << 1 << ' ' << a << '\n';
      continue;
    }
    int p = -1;
    for (int i = 2; i <= 50; i++) {
      if (qp(2, i - 2) * (a + 1) > MAX) break;
      if (qp(2, i - 2) * (a + 1) <= b && qp(2, i - 2) * (a + m) >= b) {
        p = i;
        break;
      }
    }
    if (p == -1) {
      cout << -1 << '\n';
      continue;
    }
    long long R = b - qp(2, p - 2) * (a + 1);
    long long S = a;
    for (int i = 2; i < p; i++) {
      long long now = qp(2, p - i - 1);
      long long x = min(R / now, m - 1);
      r[i] = x;
      R -= now * x;
      S += S + r[i] + 1;
    }
    S = a;
    cout << p << ' ' << S << ' ';
    for (int i = 2; i < p; i++) {
      cout << S + r[i] + 1 << ' ';
      S += S + r[i] + 1;
    }
    cout << b << '\n';
  }
  return 0;
}