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;
int T;
long long u, v;
int ans[10010];
int main() {
  scanf("%lld %lld", &u, &v);
  if (u > v || (u & 1) != (v & 1)) {
    printf("-1\n");
    return 0;
  }
  int cnt = 0;
  while ((1ll << cnt) <= v) cnt++;
  cnt--;
  for (int i = cnt; i >= 0; i--) {
    long long _2 = (1ll << i);
    ans[i] = ((u & _2) ? 1 : 0);
  }
  v -= u;
  for (int i = cnt; i >= 0; i--) {
    long long _2 = (1ll << i);
    if (v & _2) ans[i - 1] += 2;
  }
  int MAX = 0;
  for (int i = 0; i <= cnt; i++) MAX = max(MAX, ans[i]);
  printf("%d\n", MAX);
  for (int j = 1; j <= MAX; j++) {
    long long a = 0;
    for (int i = 0; i <= cnt; i++) {
      if (ans[i]) {
        a |= (1ll << i);
        ans[i]--;
      }
    }
    printf("%lld%c", a, j == MAX ? '\n' : ' ');
  }
  return 0;
}
