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;
int inf = 0x3f3f3f3f;
long long linf = 0x3f3f3f3f3f3f3f3f;
vector<int> sol;
int main() {
  long long x, s;
  cin >> x >> s;
  if (x > s) {
    puts("-1");
    return 0;
  }
  if (x == 0 && s == 0) {
    puts("0");
    return 0;
  }
  if (x == s) {
    cout << "1\n" << s << endl;
    return 0;
  }
  if ((s - x) % 2) {
    puts("-1");
    return 0;
  }
  long long a, b;
  b = a = (s - x) / 2;
  if (a + b + x != s) {
    puts("-1");
    return 0;
  }
  if (a == 0) {
    if (b + x == s && (b ^ x) == x) {
      puts("2");
      cout << b << " " << x << endl;
    } else {
      puts("-1");
    }
  } else if (b + x == (b ^ x)) {
    puts("2");
    cout << a << " " << b + x << " " << endl;
  } else {
    puts("3");
    cout << a << " " << b << " " << x << endl;
  }
  return 0;
}
