Problem: 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.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int MAXN = 100023;
vector<vector<pair<int, int> > > adj(MAXN);
pair<long long, long long> compute(long long S, long long X) {
  long long A = (S - X) / 2;
  long long a = 0, b = 0;
  long long d = 1;
  for (int i = 0; i < 61; i++) {
    long long Xi = (X & d);
    long long Ai = (A & d);
    if (Xi == 0 && Ai == 0) {
    } else if (Xi == 0 && Ai > 0) {
      a = (d | a);
      b = (d | b);
    } else if (Xi > 0 && Ai == 0) {
      a = (d | a);
    } else {
      return {-1, -1};
    }
    d = d * 2;
  }
  return {a, b};
}
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  long long u, v;
  cin >> u >> v;
  if (v == u) {
    if (v == 0) {
      cout << 0 << endl;
    } else {
      cout << 1 << endl;
      cout << v << endl;
    }
  } else if (u > v) {
    cout << -1 << endl;
  } else {
    if ((v - u) % 2) {
      cout << -1 << endl;
    } else {
      pair<long long, long long> nn = compute(v, u);
      if (nn.first == -1) {
        cout << 3 << endl;
        cout << (v - u) / 2 << " " << (v - u) / 2 << " " << u << endl;
      } else {
        cout << 2 << endl;
        cout << nn.first << " " << nn.second << endl;
      }
    }
  }
  return 0;
}