Problem: 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.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long u, v;
  cin >> u >> v;
  if (u > v || u % 2 != v % 2) {
    cout << "-1" << endl;
    return 0;
  }
  if (u == v) {
    if (u == 0)
      cout << "0" << endl;
    else {
      cout << "1" << endl;
      cout << u << endl;
    }
    return 0;
  }
  long long p = (v - u) / 2;
  if (p & u) {
    cout << "3" << endl;
    cout << u << " " << p << " " << p << endl;
  } else {
    cout << "2" << endl;
    printf("%I64d %I64d", (u ^ p), p);
  }
  return 0;
}