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 a[70];
int main() {
  long long v, u;
  scanf("%lld%lld", &v, &u);
  if (u < v || (u - v) & 1) {
    printf("-1\n");
    return 0;
  }
  int pos = 0;
  u -= v;
  while (v) {
    if (v & 1) a[pos]++;
    pos++;
    v >>= 1;
  }
  pos = 0;
  while (u) {
    if (u & 1) a[pos - 1] += 2;
    pos++;
    u >>= 1;
  }
  int ans = 0;
  for (int i = 0; i < 64; i++) ans = max(ans, a[i]);
  printf("%d\n", ans);
  while (ans--) {
    long long tmp = 0;
    for (int i = 63; i >= 0; i--) {
      tmp <<= 1;
      if (a[i]) {
        tmp++;
        a[i]--;
      }
    }
    printf("%lld ", tmp);
  }
  printf("\n");
  return 0;
}