Problem: 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.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int main() {
  long long int u, v;
  scanf("%lld%lld", &u, &v);
  if (u > v) {
    printf("-1");
    return 0;
  }
  if (u == 0 && v == 0) {
    printf("0");
    return 0;
  }
  if (u == v) {
    printf("1\n%lld", u);
    return 0;
  } else if ((v - u) % 2 == 1) {
    printf("-1");
    return 0;
  } else if ((v - u) % 2 == 0) {
    long long int a = (v - u) / 2;
    long long int b = a + u;
    if ((a ^ b) == u && a + b == v) {
      printf("2\n%lld %lld", a, b);
    } else
      printf("3\n%lld %lld %lld", a, a, u);
    return 0;
  }
}