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() {
  cin >> n >> m;
  if (n > m || (n % 2 != m % 2))
    cout << "-1\n";
  else {
    long long f = (m - n) / 2;
    if (m == 0)
      cout << "0\n";
    else if (n == m)
      cout << "1\n" << n << '\n';
    else if (!(f & n)) {
      cout << "2\n";
      cout << n + f << ' ' << f << '\n';
    } else {
      cout << "3\n";
      cout << n << ' ' << f << ' ' << f << '\n';
    }
  }
}
