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;
long long u, v;
void add(int &a, int b) {
  a += b;
  if (a >= 1000000007) a -= 1000000007;
}
void dec(int &a, int b) {
  a -= b;
  if (a < 0) a += 1000000007;
}
int pow_mod(int a, int i) {
  int s = 1;
  while (i) {
    if (i & 1) s = 1LL * s * a % 1000000007;
    a = 1LL * a * a % 1000000007;
    i >>= 1;
  }
  return s;
}
int main() {
  scanf("%lld%lld", &u, &v);
  if (u == 0 && v == 0) {
    puts("0");
    return 0;
  }
  if (u > v || (v - u) & 1) {
    puts("-1");
    return 0;
  }
  if (u == v) {
    puts("1");
    printf("%lld\n", u);
    return 0;
  }
  long long x = (v - u) / 2;
  if (!(x & u)) {
    puts("2");
    printf("%lld %lld\n", x ^ u, x);
  } else {
    puts("3");
    printf("%lld %lld %lld\n", u, x, x);
  }
  return 0;
}
