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 N = 1e5 + 5;
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  long long v, u;
  cin >> u >> v;
  if (u == v && !u) {
    return cout << "0\n", 0;
  }
  if (u > v || (u - v) & 1)
    cout << -1 << "\n";
  else {
    if (u == v)
      cout << 1 << "\n" << u << "\n";
    else {
      if ((v - u) % 2 == 0) {
        long long x = 0, y = 0;
        long long A = (v - u) / 2;
        long long dif = (v - A * 2);
        for (int i = 0; i < 60; i++) {
          if (A & (1LL << i))
            x |= (1LL << i), y |= (1LL << i);
          else if (dif & (1LL << i))
            x |= (1LL << i), dif ^= (1LL << i);
        }
        if (dif == 0) {
          cout << 2 << "\n" << x << " " << y << "\n";
        } else {
          cout << 3 << "\n";
          long long t = (v - u) / 2;
          cout << u << " " << t << " " << t << "\n";
        }
      }
    }
  }
}
