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 main() {
  long long u, v;
  scanf("%lld %lld", &u, &v);
  if (u > v) {
    puts("-1");
    return 0;
  }
  if (u == 0LL) {
    if (u == v) {
      puts("0");
    } else if (v & 1LL) {
      puts("-1");
    } else {
      puts("2");
      printf("%lld %lld\n", v / 2LL, v / 2LL);
    }
    return 0;
  }
  if (u == v) {
    cout << 1 << endl;
    cout << u << endl;
    return 0;
  }
  long long diff = v - u;
  if (diff & 1LL) {
    puts("-1");
  } else {
    long long half = diff / 2LL;
    long long bus = half;
    for (int i = 0; i < 62; i++) {
      if (u & (1LL << i)) bus |= (1LL << i);
    }
    if ((bus ^ half) == u and (bus + half == v)) {
      puts("2");
      printf("%lld %lld\n", half, bus);
    } else {
      puts("3");
      printf("%lld %lld %lld\n", half, half, u);
    }
  }
  return 0;
}
