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("Ofast")
const int mod = 1000000007;
const int inf = 1000000009;
const long long INF = 1000000000000000009;
const long long big = 1000000000000000;
const long double eps = 0.000000000000000000001;
using namespace std;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long int u, v;
  cin >> u >> v;
  if (u > v || (u & 1) != (v & 1))
    cout << -1;
  else if (u == v && v == 0)
    cout << 0;
  else if (u == v)
    cout << 1 << " " << u;
  else {
    long long int x = (v - u) / 2;
    if ((x & u) == 0)
      cout << 2 << "\n" << x << " " << x + u;
    else
      cout << 3 << "\n" << u << " " << x << " " << x;
  }
  return 0;
}
