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;
const long long int INF = (long long int)2e18 + 77;
bool parity(long long int u, long long int v) {
  if ((u % 2 == 0 && v % 2 == 0) || (u % 2 == 1 && v % 2 == 1))
    return 0;
  else
    return 1;
}
void solve() {
  long long int u, v, x;
  cin >> u >> v;
  if (u == 0 && v == 0) {
    cout << "0\n";
    return;
  }
  if (u == v) {
    cout << "1\n" << u << "\n";
    return;
  }
  if (u > v || parity(u, v))
    cout << "-1\n";
  else {
    x = (v - u) / 2;
    if (x & u) {
      cout << "3\n";
      cout << u << " " << x << " " << x << "\n";
    } else {
      cout << "2\n";
      cout << x + u << " " << x << "\n";
    }
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  solve();
  return 0;
}