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 long long md = 1e9 + 7;
const long long MX = 2e5 + 5;
const long long INF = 1e18;
const long double PI = 4 * atan((long double)1);
long long power(long long a, long long n) {
  long long res = 1;
  while (n) {
    if (n % 2) res *= a;
    a *= a;
    n /= 2;
  }
  return res;
}
long long abst(long long a) { return ((a < 0) ? (-1 * a) : (a)); }
int32_t main() {
  long long u, v;
  cin >> u >> v;
  if ((u > v) || ((u % 2) != (v % 2))) {
    cout << "-1\n";
    return 0;
  } else if (u == v) {
    if (u == 0)
      cout << 0 << "\n";
    else {
      cout << 1 << "\n";
      cout << u << "\n";
    }
    return 0;
  }
  long long x = (v - u) / 2;
  if ((u + x) == (u ^ x)) {
    cout << 2 << "\n";
    cout << (u + x) << " " << x << "\n";
  } else {
    cout << 3 << "\n";
    cout << u << " " << x << " " << x << "\n";
  }
  return 0;
}
