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 dp[55];
bool toobig(long long int a, long long int b) {
  long long int now = 1e18;
  if (now / a < b) return true;
  return false;
}
bool toobig2(long long int a, long long int b) {
  long long int now = 1e18;
  if (now - a < b) return true;
  return false;
}
int main(void) {
  int q;
  scanf("%d", &q);
  dp[0] = 1;
  dp[1] = 1;
  for (int e = 2; e <= 50; e++) dp[e] = dp[e - 1] * 2;
  while (q--) {
    long long int a, b, m;
    int err = 0;
    scanf("%lld%lld%lld", &a, &b, &m);
    if (a == b) {
      printf("1 %lld\n", a);
      continue;
    }
    for (int e = 2; e <= 50; e++) {
      if (toobig(dp[e - 1], a)) continue;
      long long int now = dp[e - 1] * a;
      if (now > b) break;
      long long int now2 = b - now;
      long long int now3 = 0, err2 = 0, now4 = 0;
      for (int p = 0; p < e - 1; p++) {
        if (toobig(dp[p], m)) {
          err2++;
          break;
        }
        if (toobig2(now3, dp[p] * m)) {
          err2++;
          break;
        }
        now3 += (dp[p] * m);
        now4 += (dp[p]);
      }
      if (err2) continue;
      if (now3 < now2) continue;
      if (now4 > now2) continue;
      now2 -= now4;
      vector<long long int> v;
      for (int p = 0; p < e - 1; p++) v.push_back(1);
      for (int p = e - 2; p >= 0; p--) {
        long long int val = min(now2 / dp[p], m - 1);
        v[(e - 2) - p] += val;
        now2 -= val * dp[p];
      }
      if (now2 > 0) continue;
      printf("%d ", e);
      printf("%lld ", a);
      vector<int> vv;
      int vis2[55];
      memset(vis2, 0, sizeof(vis2));
      for (int p = 0; p <= 50; p++) vv.push_back(dp[p]);
      for (int p = 0; p < e - 1; p++) {
        long long int ans2 = a * dp[p + 1];
        for (int q = 0; q <= p; q++) {
          ans2 += dp[vis2[q]] * v[q];
          vis2[q]++;
        }
        printf("%lld ", ans2);
      }
      err++;
      break;
    }
    if (err == 0) printf("-1");
    printf("\n");
  }
  return 0;
}
