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;
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  long long a, b;
  cin >> a >> b;
  if (b < a || (b - a) % 2)
    cout << -1 << endl;
  else if (b == a && a == 0)
    cout << 0 << endl;
  else if (b == a)
    cout << 1 << endl << a;
  else {
    long long x = 0, cnt = 1, pp = b - a;
    while (cnt <= pp) cnt *= 2;
    cnt /= 2;
    while (true) {
      if (cnt * 2 <= pp) {
        x += cnt;
        pp -= cnt * 2;
      }
      cnt /= 2;
      if (pp == 0) break;
    }
    long long tt = a ^ x;
    if (tt == a + x)
      cout << 2 << endl << a + x << ' ' << x << endl;
    else
      cout << 3 << endl << a << ' ' << x << ' ' << x << endl;
  }
}