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() {
  ios ::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  unsigned long long u, v;
  cin >> u >> v;
  if (u % 2 != v % 2 or u > v) {
    cout << -1 << '\n';
    return 0;
  }
  if (u == v) {
    if (u != 0) {
      cout << 1 << '\n';
      cout << u << '\n';
    } else
      cout << 0 << '\n';
    return 0;
  }
  int idx = 0;
  unsigned long long u1 = u, n1 = 0;
  while (u1) {
    if (u1 & 1) n1 += (1ll << idx);
    u1 >>= 1;
    idx++;
  }
  unsigned long long s = (v - n1) / 2;
  if ((s | n1) == s + n1) {
    cout << 2 << '\n';
    cout << s + n1 << " " << s << '\n';
    return 0;
  }
  cout << 3 << '\n';
  cout << 1ll * (v - u) / 2 << " " << 1ll * (v - u) / 2 << " " << u << '\n';
  return 0;
}
