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;
#pragma GCC target("avx2")
#pragma GCC optimization("O3")
#pragma GCC optimization("unroll-loops")
struct greateri {
  template <class T>
  bool operator()(T const &a, T const &b) const {
    return a > b;
  }
};
void setIO(string s) {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  freopen((s + ".in").c_str(), "r", stdin);
  freopen((s + ".out").c_str(), "w", stdout);
}
int main() {
  long long u, v;
  cin >> u >> v;
  if (u > v || (abs(u - v)) % 2)
    cout << -1;
  else if (u == 0 && v == 0)
    cout << 0;
  else if (u == v) {
    cout << 1 << '\n';
    cout << u;
  } else {
    long long a = (v - u) / 2;
    bool ka = 1;
    long long ans1 = 0, ans2 = 0;
    for (int i = 0; i <= 64; ++i) {
      if ((u & ((long long)(pow(2, i)))) && !(a & ((long long)(pow(2, i))))) {
        ans1 |= ((long long)(pow(2, i)));
      }
      if (!(u & ((long long)(pow(2, i)))) && (a & ((long long)(pow(2, i))))) {
        ans1 |= ((long long)(pow(2, i)));
        ans2 |= ((long long)(pow(2, i)));
      }
      if ((u & ((long long)(pow(2, i)))) && (a & ((long long)(pow(2, i))))) {
        ka = 0;
        break;
      }
    }
    if (ka) {
      cout << 2 << '\n';
      cout << ans1 << " " << ans2;
    } else {
      cout << 3 << '\n';
      cout << u << " " << (v - u) / 2 << " " << (v - u) / 2;
    }
  }
  return 0;
}
