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 power(long long a, long long n, long long mod) {
  long long res = 1;
  while (n > 0) {
    if (n & 1) {
      res = (res * a) % mod;
    }
    a = (a * a) % mod;
    n = n / 2;
  }
  return res;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int t = 1;
  while (t--) {
    unsigned long long u, v;
    cin >> u >> v;
    if (u > v || (v - u) % 2) {
      cout << "-1\n";
    } else if (u == v) {
      if (!v) {
        cout << "0\n";
      } else {
        cout << "1\n" << u << "\n";
      }
    } else {
      unsigned long long _and = (v - u) / 2;
      if (u & _and) {
        cout << "3\n";
        cout << _and << " " << _and << " " << u << "\n";
      } else {
        cout << "2\n";
        cout << (u ^ _and) << " " << _and << "\n";
      }
    }
  }
}
