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;
void prexit() {
  printf("-1\n");
  exit(0);
}
int getBit(long long x, int pos) {
  if (x & (1LL << pos)) return 1;
  return 0;
}
int main() {
  long long u, v;
  scanf("%lld %lld", &u, &v);
  if (u > v || (v - u) % 2) prexit();
  if (u + v == 0) {
    printf("0\n");
    return 0;
  }
  if (u == v) {
    printf("1\n%lld\n", u);
    return 0;
  }
  int c = 0;
  long long a = 0, b = 0;
  bool flag = true;
  for (int i = 0; i < 62; i++) {
    if (getBit(u, i)) {
      b |= (1LL << i);
      if (getBit(v, i) != (c ^ 1)) {
        flag = false;
        break;
      }
    } else {
      if (getBit(v, i) != c) {
        flag = false;
        break;
      }
      if (getBit(u, i + 1) == getBit(v, i + 1))
        c = 0;
      else
        a |= (1LL << i), b |= (1LL << i), c = 1;
    }
  }
  if (flag) {
    printf("2\n%lld %lld\n", a, b);
    return 0;
  }
  printf("3\n%lld %lld %lld\n", u, (v - u) / 2, (v - u) / 2);
  return 0;
}
