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>
#pragma GCC optimize("O3")
#pragma GCC tsmthget("avx2")
using namespace std;
const long long one = 1;
const long long INF = 2e18;
const long long MOD = 1e9 + 7;
const long long N = 5e5 + 10;
const long long nul = 0;
signed main() {
  cin.tie(0);
  cout.tie(0);
  ios_base::sync_with_stdio(0);
  long long v, u;
  cin >> u >> v;
  if (u > v) {
    cout << -1 << endl;
  } else if (u == v && u != 0) {
    cout << 1 << endl << u << endl;
  } else if (u == v && v == 0) {
    cout << 0 << endl;
  } else {
    long long d = v - u;
    if ((d ^ u) == u) {
      cout << 2 << endl;
      cout << d << " " << u << endl;
    } else if ((u ^ d) + d == v) {
      cout << 2 << endl;
      cout << (u ^ d) << " " << d << endl;
    } else if (((d / 2) ^ (v - d / 2)) == u) {
      cout << 2 << endl;
      cout << (d / 2) << " " << (v - d / 2) << endl;
    } else {
      if (d % 2 == 1) {
        cout << -1 << endl;
      } else {
        cout << 3 << endl;
        cout << d / 2 << " " << d / 2 << " " << u << endl;
      }
    }
  }
}
