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;
long long a, b, lg[61], s1[61], ans[110000];
int main() {
  lg[0] = 1;
  for (long long i = 1; i <= 60; i++) lg[i] = lg[i - 1] << 1;
  scanf("%lld%lld", &a, &b);
  for (long long i = 60; i >= 0; i--)
    if (lg[i] & a) s1[i] = 1;
  b -= a;
  if (b < 0 || b & 1) {
    printf("-1\n");
    return 0;
  }
  long long k = 0;
  for (long long i = 60; i >= 0; i--) {
    long long c = b / lg[i] / 2 * 2;
    s1[i] += c;
    b -= lg[i] * c;
    k = max(k, s1[i]);
  }
  printf("%lld\n", k);
  for (long long i = 1; i <= k; i++) {
    long long ans = 0;
    for (long long i = 60; i >= 0; i--) {
      if (s1[i]) {
        s1[i] -= 1;
        ans += lg[i];
      }
    }
    printf("%lld ", ans);
  }
  return 0;
}