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;
const int INF = 0x3f3f3f3f;
mt19937 gen(chrono::high_resolution_clock::now().time_since_epoch().count());
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  long long u, v;
  cin >> u >> v;
  if ((v - u) % 2 || u > v)
    cout << -1;
  else if (!u && !v)
    cout << 0;
  else if (u == v)
    cout << "1\n" << u;
  else {
    long long p = (v - u) / 2;
    if (p + (u ^ p) == v)
      cout << "2\n" << p << ' ' << (u ^ p);
    else
      cout << "3\n" << u << ' ' << p << ' ' << p;
  }
  return 0;
}
