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() {
  long long int t;
  t = 1;
  while (t--) {
    long long int u, v, x;
    cin >> u >> v;
    if (u > v)
      cout << -1 << endl;
    else if ((u % 2 != 0 && v % 2 == 0) || (u % 2 == 0 && v % 2 != 0))
      cout << -1 << endl;
    else if (u == v && u == 0)
      cout << 0 << endl;
    else if (u == v) {
      cout << 1 << endl;
      cout << u << endl;
    } else {
      x = (v - u) / 2;
      long long int ans = u & x;
      if (ans == 0) {
        cout << 2 << " " << endl;
        cout << u + x << " " << x << endl;
      } else {
        cout << 3 << " " << endl;
        cout << u << " " << x << " " << x << endl;
      }
    }
  }
}
