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() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int u, v;
  cin >> u >> v;
  if (u == 0 && v == 0) {
    cout << 0 << endl;
    return 0;
  }
  if (u == v) {
    cout << 1 << endl;
    cout << v << endl;
    return 0;
  }
  if (u > v || (v - u) % 2) {
    cout << -1 << endl;
    return 0;
  }
  long long int x = (v - u) / 2;
  if ((x & u)) {
    cout << 3 << endl;
    cout << u << " " << x << " " << x << endl;
    return 0;
  }
  long long int z1 = 0, z2 = 0;
  for (long long int i = 0; i <= 60; i++) {
    long long int s1 = u % 2, s2 = x % 2;
    if (s1 == 0 && s2 == 1) {
      z1 += (1ll << i);
      z2 += (1ll << i);
    }
    if (s1 == 1 && s2 == 0) {
      z1 += (1ll << i);
    }
    u /= 2;
    x /= 2;
  }
  cout << 2 << endl;
  cout << z1 << " " << z2 << endl;
  return 0;
}
