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 long long inf = 1e15 + 1;
const int N = 1000002;
long long gcd(long long x, long long y) {
  if (y == 0) return x;
  return gcd(y, x % y);
}
long long power(long long A, long long B, long long mod = 1000000007) {
  long long r = 1;
  while (B > 0) {
    if (B & 1LL) r = (r * A) % mod;
    A = (A * A) % mod;
    B >>= 1;
  }
  return r;
}
struct set_comp {
  bool operator()(const int& lhs, const int& rhs) { return lhs < rhs; }
};
vector<long long> solve(long long a, long long b, long long k, long long m) {
  vector<long long> v;
  if (k == 1) {
    if (a == b) v.push_back(a);
    return v;
  } else if (k == 2) {
    if (b - a <= m and b - a >= 1) {
      v.push_back(a);
      v.push_back(b);
    }
    return v;
  }
  if ((1LL << (k - 2)) > inf / (a + 1)) {
    vector<long long> v;
    return v;
  }
  long long term = (a + 1LL) * (1LL << (k - 2LL));
  if (term > b) {
    vector<long long> v;
    return v;
  }
  long long rem = b - term;
  v.push_back(a);
  long long sum = a;
  for (long long i = 2; i <= k - 1; i++) {
    long long a_i = sum + 1;
    v.push_back(a_i);
    sum = sum + a_i;
  }
  v.push_back(b);
  long long coin = (1LL << (k - 3));
  sum = 0;
  for (long long i = 2; i <= k - 2; i++) {
    long long ext;
    if (rem / coin > m - 1) {
      ext = sum + m - 1;
      rem = rem - (m - 1) * coin;
    } else {
      ext = sum + rem / coin;
      rem = rem - (rem / coin) * coin;
    }
    v[i - 1] = v[i - 1] + ext;
    sum = sum + ext;
    coin >>= 1LL;
  }
  if (rem <= 2 * m - 2) {
    long long ext = sum + rem / 2;
    v[k - 2] = v[k - 2] + ext;
    sum = sum + ext;
    rem = rem - rem / 2;
    ext = sum + rem;
    return v;
  } else {
    vector<long long> v;
    return v;
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(NULL), cout.tie(NULL);
  long long q;
  cin >> q;
  while (q--) {
    long long a, b, m, k;
    cin >> a >> b >> m;
    for (k = 1; k <= 50; k++) {
      vector<long long> v = solve(a, b, k, m);
      if (v.size() > 0) {
        cout << k << " ";
        for (long long i = 0; i < v.size(); i++) cout << v[i] << " ";
        cout << endl;
        break;
      }
    }
    if (k == 51) cout << -1 << endl;
  }
  return 0;
}
