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 arr[5];
int main() {
  long long u, v;
  scanf("%lld %lld", &u, &v);
  long long a = v - u;
  if (a < 0 || a % 2) {
    printf("-1\n");
    return 0;
  }
  a = a >> 1;
  long long cou = 1;
  while (a || u) {
    if (a % 2 && u % 2) {
      arr[1] += cou;
      arr[2] += cou;
      arr[3] += cou;
    } else if (a % 2 && !(u % 2)) {
      arr[1] += cou;
      arr[2] += cou;
    } else if (!(a % 2) && u % 2) {
      arr[1] += cou;
    }
    a /= 2;
    u /= 2;
    cou *= 2;
  }
  if (arr[1] == 0) {
    printf("0\n");
  } else if (arr[2] == 0) {
    printf("1\n");
    printf("%lld\n", arr[1]);
  } else if (arr[3] == 0) {
    printf("2\n");
    printf("%lld %lld\n", arr[1], arr[2]);
  } else {
    printf("3\n");
    printf("%lld %lld %lld\n", arr[1], arr[2], arr[3]);
  }
  return 0;
}
