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;
using LL = long long;
const int MAXN(2e5 + 100);
int n;
LL a, b, m;
inline int read() {
  int f = 1, x = 0;
  char ch;
  do {
    ch = getchar();
    if (ch == '-') f = -1;
  } while (ch < '0' || ch > '9');
  do {
    x = x * 10 + ch - '0';
    ch = getchar();
  } while (ch >= '0' && ch <= '9');
  return f * x;
}
LL binsearch(LL c, LL target) {
  LL l = 1, r = m;
  while (l < r) {
    LL mid = (l + r) >> 1;
    if (c * mid <= target)
      l = mid + 1;
    else
      r = mid;
  }
  return l - 1;
}
void init() { scanf("%I64d%I64d%I64d", &a, &b, &m); }
void solve() {
  if (a == b) {
    printf("1 %I64d\n", a);
    return;
  }
  LL low = a + 1, up = a + m;
  n = 2;
  while (n <= 50 && low <= b) {
    if (low <= b && b <= up) break;
    low <<= 1;
    up <<= 1;
    n++;
  }
  if (n > 50 || low > b)
    printf("-1\n");
  else {
    LL rem = b - (1LL << (n - 2)) * a;
    for (int i = n - 3; i >= 0; i--) rem -= (1LL << i);
    rem -= 1LL;
    vector<LL> r;
    for (int i = n - 3; i >= 0; i--) {
      LL tmp = binsearch(1LL << i, rem);
      rem -= (1LL << i) * tmp;
      r.push_back(tmp + 1);
    }
    r.push_back(rem + 1);
    printf("%d %I64d ", n, a);
    LL x = a + r[0];
    for (int i = 2; i <= n; i++) {
      printf("%I64d%c", x, i < n ? ' ' : '\n');
      if (i < n) x = 2 * x + r[i - 1] - r[i - 2];
    }
  }
}
int main() {
  int t = read();
  while (t--) {
    init();
    solve();
  }
}
