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;
long long Move[8][2] = {{0, 1}, {1, 0},  {0, -1}, {-1, 0},
                        {1, 1}, {-1, 1}, {1, -1}, {-1, -1}};
void sync() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
}
int main() {
  unsigned long long a, b;
  cin >> a >> b;
  if ((a % 2) != (b % 2) || a > b) {
    cout << -1;
    return 0;
  }
  if (a == 0 && b == 0) {
    cout << 0;
    return 0;
  }
  if (a == b) {
    cout << 1 << "\n" << a;
    return 0;
  }
  unsigned long long r = b - a;
  r /= 2;
  if (a & r) {
    cout << 3 << "\n" << a << " " << r << " " << r;
  } else
    cout << 2 << "\n" << (unsigned long long)(a ^ r) << " " << r;
}