Given 2 integers u and v, find the shortest array such that [bitwise-xor](https://en.wikipedia.org/wiki/Bitwise_operation#XOR) of its elements is u, and the sum of its elements is v.

Input

The only line contains 2 integers u and v (0 ≤ u,v ≤ 10^{18}).

Output

If there's no array that satisfies the condition, print "-1". Otherwise:

The first line should contain one integer, n, representing the length of the desired array. The next line should contain n positive integers, the array itself. If there are multiple possible answers, print any.

Examples

Input


2 4


Output


2
3 1

Input


1 3


Output


3
1 1 1

Input


8 5


Output


-1

Input


0 0


Output


0

Note

In the first sample, 3⊕ 1 = 2 and 3 + 1 = 4. There is no valid array of smaller length.

Notice that in the fourth sample the array is empty.
#include <bits/stdc++.h>
using namespace std;
int main(int argc, char **argv) {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long u, v;
  cin >> u >> v;
  if (u > v) {
    cout << -1 << endl;
    return 0;
  }
  if (u == v && u == 0) {
    cout << 0 << endl;
    return 0;
  }
  vector<int> nu(60);
  long long te = u;
  int pos = 0;
  while (te) nu[59 - pos] = te & 1, te /= 2, pos++;
  int prev = 0;
  for (int i = 59; i >= 0; i--) {
    if (((nu[i] + (prev >> 1)) & 1) == (((1LL << (59 - i)) & v) >> (59 - i))) {
      prev = nu[i] + (prev >> 1);
      continue;
    }
    if (i == 59) {
      cout << -1 << endl;
      return 0;
    }
    nu[i + 1] += 2;
    prev = nu[i] + (prev >> 1) + 1;
  }
  int n = *max_element(nu.begin(), nu.end());
  vector<long long> ans(n, 0);
  for (int i = 0; i < 60; i++) {
    for (int j = 0; j < nu[i]; j++) {
      ans[j] += (1LL << (59 - i));
    }
  }
  cout << n << endl;
  for (auto i : ans) cout << i << " ";
  cout << "\n";
  cout << endl;
  return 0;
}
