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 double PI = acos(-1.0);
mt19937 rng(
    (unsigned int)chrono::steady_clock::now().time_since_epoch().count());
long long power(long long a, long long b) {
  long long result = 1;
  while (b > 0) {
    if (b % 2 == 1) {
      result *= a;
    }
    a *= a;
    b /= 2;
  }
  return result;
}
long long gcd(long long x, long long y) {
  long long r;
  while (y != 0 && (r = x % y) != 0) {
    x = y;
    y = r;
  }
  return y == 0 ? x : y;
}
long long countSetBits(long long x) {
  long long Count = 0;
  while (x > 0) {
    if (x & 1) Count++;
    x = x >> 1;
  }
  return Count;
}
bool isPerfectSquare(long long n) {
  long long sr = sqrt(n);
  if (sr * sr == n)
    return true;
  else
    return false;
}
long long mod(long long x, long long M) { return ((x % M + M) % M); }
long long add(long long a, long long b, long long M) {
  return mod(mod(a, M) + mod(b, M), M);
}
long long mul(long long a, long long b, long long M) {
  return mod(mod(a, M) * mod(b, M), M);
}
long long powerM(long long a, long long b, long long M) {
  long long res = 1ll;
  while (b) {
    if (b % 2ll == 1ll) {
      res = mul(a, res, M);
    }
    a = mul(a, a, M);
    b /= 2ll;
  }
  return res;
}
long long mod_inv(long long a, long long m) {
  long long g = m, r = a, x = 0, y = 1;
  while (r != 0) {
    long long q = g / r;
    g %= r;
    swap(g, r);
    x -= q * y;
    swap(x, y);
  }
  return mod(x, m);
}
long long nCr(long long n, long long k) {
  if (n < k) return 0;
  if (k == 0) return 1;
  long long res = 1;
  if (k > n - k) k = n - k;
  for (long long i = 0; i < k; ++i) {
    res *= (n - i);
    res /= (i + 1);
  }
  return res;
}
long long modInverse(long long n, long long M) { return powerM(n, M - 2, M); }
vector<long long> factu;
void factcalc(long long n, long long M) {
  factu.resize(n + 1);
  factu[0] = 1;
  for (long long i = 1; i <= n; i++) {
    factu[i] = mul(factu[i - 1], i, M);
  }
}
long long nCrM(long long n, long long r, long long M) {
  if (n < r) return 0;
  if (r == 0) return 1;
  return mul(mul(factu[n], modInverse(factu[r], M), M),
             modInverse(factu[n - r], M), M);
}
long long rand_int(long long l, long long r) {
  return uniform_int_distribution<long long>(l, r)(rng);
}
void solve() {
  long long a, b, m;
  cin >> a >> b >> m;
  if (a == b) {
    cout << "1 " << a << endl;
    return;
  } else if (a + m >= b) {
    cout << "2 " << a << " " << b << endl;
    return;
  }
  for (long long k = 2; k <= 50; k++) {
    vector<long long> v(k + 1);
    vector<long long> psum(k + 1);
    v[1] = a;
    psum[1] = a;
    long long f = 0;
    for (long long i = 2; i <= k; i++) {
      v[i] = psum[i - 1] + 1;
      if (psum[i - 1] >= 1e16) {
        f = 1;
        break;
      }
      psum[i] = psum[i - 1] + v[i];
    }
    if (f) {
      continue;
    }
    if (v[k] <= b) {
      long long curr = b - v[k];
      vector<long long> carry(k + 1);
      for (long long i = 2; i < k; i++) {
        long long fuel = m - 1;
        if (curr > (fuel) * (1ll << (k - i - 1))) {
          curr -= (fuel) * (1ll << (k - i - 1));
          carry[i] += fuel;
        } else {
          long long gpl = curr / ((1ll << (k - i - 1)));
          curr -= gpl * (1ll << (k - i - 1));
          carry[i] += gpl;
        }
      }
      if (curr <= m - 1) {
        carry[k] += curr;
        curr = 0;
      }
      if (curr != 0) {
        continue;
      }
      for (long long i = 2; i <= k; i++) {
        long long cnt = 1;
        long long pos = carry[i];
        v[i] += pos * cnt;
        for (long long j = i + 1; j <= k; j++) {
          v[j] += pos * cnt;
          cnt *= 2;
        }
      }
      cout << k << " ";
      for (long long i = 1; i <= k; i++) {
        cout << v[i] << " ";
      }
      cout << endl;
      return;
    } else {
      continue;
    }
  }
  cout << -1 << endl;
}
int main() {
  ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  cout << fixed << setprecision(20);
  long long t = 0;
  cin >> t;
  while (t--) solve();
  return 0;
}
