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;
const int N = 300010;
int gi() {
  int f = 0, op = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') op = -op;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') f = f * 10 + ch - 48, ch = getchar();
  return f * op;
}
long long n, m;
int t;
long long ans[4];
int cnt = 2;
int main() {
  scanf("%lld%lld", &n, &m);
  if (n > m) {
    puts("-1");
    return 0;
  }
  if (!n && !m) {
    puts("0");
    return 0;
  }
  m -= n;
  if (m & 1) {
    puts("-1");
    return 0;
  }
  ans[1] = n;
  if (!m) {
    printf("1\n%lld\n", n);
    return 0;
  }
  for (int i = 61; i >= 0; i--) {
    long long tmp = 1ll << i;
    if (m < tmp * 2) continue;
    m -= tmp * 2;
    if (ans[1] & tmp) {
      cnt = 3;
      ans[3] |= tmp;
      ans[2] |= tmp;
    } else
      ans[2] |= tmp, ans[1] |= tmp;
    if (!m) break;
  }
  printf("%d\n", cnt);
  for (int i = 1; i <= cnt; i++) printf("%lld ", ans[i]);
  return 0;
}
