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() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  long long u, v;
  cin >> u >> v;
  if (!u && !v) {
    cout << 0;
    return 0;
  }
  if (u == v) {
    cout << "1\n" << u;
    return 0;
  }
  if (u & 1 ^ v & 1 || u > v) {
    cout << -1;
    return 0;
  }
  long long _and = (v - u) / 2;
  long long c = u + _and, d = _and;
  if ((c ^ d) == u) {
    cout << "2\n" << c << " " << d << "\n";
    return 0;
  }
  cout << "3\n" << (v - u) / 2 << " " << (v - u) / 2 << " " << u;
}
