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;
using ll = long long int;
using ui = unsigned int;
using pii = pair<int, int>;
using pll = pair<int, int>;
struct fast_ios {
  fast_ios() {
    ios::sync_with_stdio(false);
    cin.tie(NULL);
  }
} fast_ios_;
ll u, v, a;
int solve() {
  cin >> u >> v;
  if (u > v || (v - u) & 1) return puts("-1"), 0;
  if (u + v == 0) return puts("0"), 0;
  if (u == v) return printf("1\n%lld\n", u), 0;
  a = (v - u) / 2;
  if (u & a) return printf("3\n%lld %lld %lld\n", u, a, a), 0;
  return printf("2\n%lld %lld\n", u + a, a), 0;
}
int main() {
  solve();
  return 0;
}
