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;
int main() {
  int tt;
  tt = 1;
  while (tt--) {
    long long x, s;
    cin >> x >> s;
    if (s < x || (s % 2) != (x % 2)) {
      cout << -1 << endl;
    } else if (s == 0) {
      cout << 0 << endl;
    } else if (s == x) {
      cout << 1 << endl << s << endl;
    } else {
      long long c = x ^ s;
      long long sorg = s;
      long long xorg = x;
      s = s - (c / 2);
      int n = 2;
      if ((s % 2) == 0 && (x % 2) == 1) {
        n++;
        s--;
        x--;
      }
      long long a = 0, b = 0;
      int pos = 0;
      while (s || x) {
        if (x % 2 == 1) {
          a += (pow(2, pos));
        } else {
          if (s % 2 == 1) {
            a += (pow(2, pos));
            b += (pow(2, pos));
          }
        }
        pos++;
        s /= 2;
        x /= 2;
      }
      if (a + b == sorg) {
        cout << 2 << endl << a << ' ' << b << endl;
      } else {
        sorg = (sorg - xorg) / 2;
        cout << 3 << endl << xorg << ' ' << sorg << ' ' << sorg << endl;
      }
    }
  }
}