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 INF = 1e9;
const long long INFll = 2e18;
const long long base1 = 239;
const long long base2 = 653;
const long long mod = 1e9 + 7;
const double eps = 1e-6;
const long long len = 140;
void solve() {
  long long u, v;
  cin >> u >> v;
  if (u == 0 && v == 0) {
    cout << 0;
    return;
  }
  if (u == v) {
    cout << "1\n" << u;
    return;
  }
  if (u > v || (u + v) % 2) {
    cout << -1;
    return;
  }
  long long a0 = u, a1 = 0;
  for (long long i = 59; i > -1; i--) {
    if (u & (1ll << i)) continue;
    if (a0 + a1 + (1ll << (i + 1)) <= v) {
      a0 += (1ll << i);
      a1 += (1ll << i);
    }
  }
  if (a0 + a1 == v) {
    cout << "2\n";
    cout << a0 << ' ' << a1;
    return;
  }
  cout << "3\n";
  cout << (v - u) / 2 << ' ' << (v - u) / 2 << ' ' << u;
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  solve();
}
