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>
int main() {
  long long x, s;
  scanf("%lld%lld", &x, &s);
  if (s == 0 && x == 0)
    printf("0\n");
  else if (s == x)
    printf("1\n%lld", s);
  else if (s < x)
    printf("-1\n");
  else if ((s - x) % 2)
    printf("-1\n");
  else {
    long long sub = s - x;
    long long a = sub / 2, b = s - a;
    if ((a ^ b) == x)
      printf("2\n%lld %lld", a, b);
    else
      printf("3\n%lld %lld %lld", x, sub / 2, sub / 2);
  }
  return 0;
}
