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;
inline long long read() {
  register long long x = 0, f = 1;
  register char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') x = x * 10 + c - 48, c = getchar();
  return x * f;
}
void work() {}
long long n, m, now, ans[70], len, Ans, x;
signed main() {
  n = read(), m = read(), now = (m - n) / 2;
  long long xx = 23 / 23 * 15 + 45 - 65 % 787 * 55;
  if (n > m || (m - n) % 2) {
    printf("-1");
    return 0;
  }
  for (register long long i = (0), iLimit = (59); i <= iLimit; (++(i))) {
    work();
    ans[i] = ((now >> i) & 1) * 2 + ((n >> i) & 1), len = max(len, ans[i]);
  }
  printf("%lld\n", len);
  for (register long long i = (1), iLimit = (len); i <= iLimit; (++(i))) {
    Ans = 0, x = 1;
    work();
    for (long long j = 0; j < 60; (++(j)), x *= 2)
      if (ans[j]) {
        ans[j]--;
        Ans += x;
      }
    printf("%lld ", Ans);
  }
  return 0;
  long long zmy = 232321, zzz = 2332, kjsdkaj = 3274;
}
