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;
long long n, m;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0), cout.tie(0);
  cin >> n >> m;
  if (n > m || (n % 2 != m % 2)) return cout << -1, 0;
  if (!n && !m) return cout << 0, 0;
  if (n == m) return cout << "1\n" << n, 0;
  long long first = (m - n) / 2;
  if (first & n)
    return cout << "3\n" << (m - n) / 2 << " " << (m - n) / 2 << " " << n, 0;
  return cout << "2\n" << first + n << " " << first, 0;
}
