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 int xor_, sum;
  cin >> xor_ >> sum;
  long long int diff = sum - xor_;
  if (diff == 0) {
    if (sum == 0) {
      cout << 0 << endl;
      return 0;
    }
    cout << 1 << endl;
    cout << sum << endl;
    return 0;
  }
  if (diff % 2 != 0 || diff < 0) {
    cout << "-1" << endl;
    return 0;
  }
  long long int and_ = diff / 2;
  bool res = true;
  long long int a = 0, b = 0;
  for (int i = 0; i < 63 && res; i++) {
    long long int and_bit = ((and_ >> i) & 1ll);
    long long int xor_bit = ((xor_ >> i) & 1ll);
    if ((xor_bit == 1 && and_bit == 1)) {
      res = false;
      a = 0;
      b = 0;
    } else if (xor_bit == 1 && and_bit == 0) {
      a = a | (1ll << i);
    } else if (xor_bit == 0 && and_bit == 1) {
      a = a | (1ll << i);
      b = b | (1ll << i);
    }
  }
  if (res) {
    cout << "2" << endl;
    cout << a << " " << b << endl;
    return 0;
  }
  a = xor_;
  b = diff / 2;
  long long int c = diff / 2;
  cout << "3" << endl;
  cout << a << " " << b << " " << c << endl;
  return 0;
}
