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;
using ll = long long;
using ii = pair<int, int>;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int q;
  cin >> q;
  for (int cas = 0; cas < q; ++cas) {
    ll a, b, m;
    cin >> a >> b >> m;
    if (a == b) {
      cout << 1 << " " << a << '\n';
      continue;
    }
    if (b - a <= m) {
      cout << 2 << " " << a << " " << b << '\n';
      continue;
    }
    bool poss = false;
    for (ll ans = 3; ans <= 50; ++ans) {
      vector<ll> sum(ans + 1, 0);
      sum[1] = a;
      sum[2] = a + 1;
      bool loc = true;
      for (int i = 3; i <= ans; ++i) {
        for (int j = 1; j < i; ++j) {
          sum[i] += sum[j];
          if (sum[i] > b) {
            loc = false;
            break;
          }
        }
        sum[i]++;
        if (sum[i] > b) loc = false;
        if (!loc) break;
      }
      if (!loc) continue;
      vector<ll> v(ans + 1, 1);
      for (int i = 2; i <= ans; ++i) {
        ll l = 0, r = m;
        for (int j = 0; j < 50; ++j) {
          ll mm = (l + r) / 2;
          if (i == ans) {
            if (sum[ans] + mm > b)
              r = mm;
            else
              l = mm;
          } else {
            if (sum[ans] + mm * (1LL << (ans - (i + 1LL))) > b)
              r = mm;
            else
              l = mm;
          }
        }
        v[i] += l;
        if (i == ans)
          sum[ans] += l;
        else
          sum[ans] += (l * (1LL << (ans - (i + 1LL))));
      }
      if (sum[ans] > b) break;
      if (sum[ans] == b) {
        cout << ans << " ";
        ll curr = a;
        vector<ll> res(ans + 1, 0);
        res[1] = a;
        for (int i = 2; i <= ans; ++i) {
          for (int j = 1; j < i; ++j) res[i] += res[j];
          res[i] += v[i];
        }
        for (int i = 1; i <= ans; ++i) cout << res[i] << " ";
        cout << '\n';
        poss = true;
        break;
      }
    }
    if (!poss) cout << -1 << '\n';
  }
  return 0;
}