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;
const double pi = 3.141592653589;
long long int mod = 1000000007;
int main() {
  int t = 1;
  while (t--) {
    long long int u, v, x;
    cin >> u >> v;
    x = (v - u) / 2;
    if (u + v == 0) {
      cout << 0;
    } else if (u == v) {
      cout << 1 << endl;
      cout << v << endl;
    } else if (u > v) {
      cout << -1;
    } else if ((v - u) % 2 == 1) {
      cout << -1;
    } else if (u == 0) {
      cout << 2 << endl;
      cout << v / 2 << " " << v / 2;
    } else if ((u & x) == 0) {
      cout << 2 << endl;
      cout << x << " " << u + x;
    } else {
      cout << 3 << endl;
      cout << (v - u) / 2 << " " << (v - u) / 2 << " " << u;
    }
  }
}
