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 a, b, num;
vector<long long> v;
int main() {
  cin >> a >> b;
  if (a > b) {
    cout << -1;
  } else if (a == 0 && b == 0) {
    cout << 0;
  } else {
    long long c = b - a;
    if (c % 2)
      cout << -1;
    else {
      if (((a + c / 2) ^ (c / 2)) == a) {
        if (a + c / 2 != 0) v.push_back(a + c / 2);
        if (c / 2 != 0) v.push_back(c / 2);
        cout << v.size() << endl;
        for (int i = 0; i < v.size(); i++) {
          cout << v[i];
          if (i != v.size() - 1) cout << " ";
        }
      } else {
        if (a != 0) v.push_back(a);
        if (c / 2 != 0) v.push_back(c / 2), v.push_back(c / 2);
        cout << v.size() << endl;
        for (int i = 0; i < v.size(); i++) {
          cout << v[i];
          if (i != v.size() - 1) cout << " ";
        }
      }
    }
  }
  return 0;
}
