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;
template <typename T>
void out(T x) {
  cout << x << endl;
  exit(0);
}
const int maxn = 1e6 + 5;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long s, x;
  cin >> x >> s;
  if (x > s) out(-1);
  if (x == s) {
    if (x == 0) out(0);
    cout << 1 << endl;
    cout << x << endl;
    return 0;
  }
  if ((s - x) & 1) {
    out(-1);
  }
  long long y = (s - x) / 2;
  assert((y + y + x) == s && ((y ^ y ^ x) == x));
  if ((x & y) == 0) {
    cout << 2 << endl;
    cout << (x | y) << " " << y << endl;
  } else {
    cout << 3 << endl;
    cout << x << " " << y << " " << y << endl;
  }
  return 0;
}
