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 = 1e5 + 2;
const long long mod = 1e9 + 7;
long long powermodm(long long x, long long n, long long M) {
  long long result = 1;
  while (n > 0) {
    if (n % 2 == 1) result = (result * x) % M;
    x = (x * x) % M;
    n = n / 2;
  }
  return result;
}
long long power(long long x, long long y) {
  long long z = 1;
  while (y) {
    if (y % 2 == 1) z = (z * x);
    y /= 2;
    x = (x * x);
  }
  return z;
}
long long gcd(long long a, long long b) {
  if (a == 0) return b;
  return gcd(b % a, a);
}
long long lcm(long long a, long long b) {
  return (max(a, b) / gcd(a, b)) * min(a, b);
}
int main() {
  long long u, v;
  cin >> u >> v;
  if (u > v)
    cout << -1;
  else if (u == v && u == 0)
    cout << 0;
  else if (u == v)
    cout << 1 << endl << u;
  else if ((v - u) % 2)
    cout << -1;
  else if (((u + (v - u) / 2) ^ ((v - u) / 2)) == u)
    cout << 2 << endl << u + (v - u) / 2 << " " << (v - u) / 2;
  else
    cout << 3 << endl << u << " " << (v - u) / 2 << " " << (v - u) / 2;
  return 0;
}
