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 xr, sm;
  cin >> xr >> sm;
  if (xr > sm || xr % 2 != sm % 2) {
    cout << -1;
    return 0;
  }
  if (xr == 0 && sm == 0) {
    cout << 0;
    return 0;
  }
  if (xr == 0) {
    cout << 2 << endl;
    cout << sm / 2 << " " << sm / 2 << endl;
    return 0;
  }
  if (sm == xr) {
    cout << 1 << endl;
    cout << sm;
    return 0;
  }
  long long am = (sm + xr) / 2;
  long long sc = sm - am;
  if (sc > 0 && (am ^ sc) == xr) {
    cout << 2 << endl << am << " " << sc;
    return 0;
  }
  cout << 3 << endl;
  cout << xr << " " << (sm - xr) / 2 << " " << (sm - xr) / 2;
}
