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>
const long long int mod = 1e9 + 7;
using namespace std;
long long int power(long long int x, long long int n) {
  long long int ans = 1;
  while (n > 0) {
    if (n & 1) ans = ans * x;
    n = n >> 1;
    x = x * x;
  }
  return ans;
}
long long int gcd(long long int x, long long int y) {
  if (y == 0) return x;
  return gcd(y, x % y);
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  long long int u, v;
  cin >> u >> v;
  if ((u > v) || ((u & 1) != (v & 1))) {
    cout << "-1" << endl;
    return 0;
  }
  if (u == v) {
    if (u == 0)
      cout << "0" << endl;
    else
      cout << "1" << endl << u << endl;
  } else {
    long long int x = (v - u) / 2;
    if (u & x)
      cout << "3" << endl << u << " " << x << " " << x << endl;
    else
      cout << "2" << endl << (u ^ x) << " " << x << endl;
  }
  return 0;
}
