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;
long long getBit(long long num, long long idx) {
  return (((num >> idx) & 1LL) == 1);
}
long long setBit(long long num, long long idx) { return (num | (1LL << idx)); }
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long n, m;
  cin >> n >> m;
  long long rem = m - n;
  if (n > m || rem & 1) return cout << -1, 0;
  vector<long long> v;
  v.push_back(n);
  v.push_back(0);
  v.push_back(0);
  for (int i = 1; i <= 60; ++i) {
    int rem_bit = getBit(rem, i);
    if (rem_bit) {
      int f_bit = getBit(v[0], i - 1);
      if (f_bit) {
        v[1] = setBit(v[1], i - 1);
        v[2] = setBit(v[2], i - 1);
      } else {
        v[0] = setBit(v[0], i - 1);
        v[1] = setBit(v[1], i - 1);
      }
    }
  }
  vector<long long> out;
  for (int i = 0; i < 3; ++i)
    if (v[i]) out.push_back(v[i]);
  cout << out.size() << '\n';
  for (auto it : out) cout << it << ' ';
  return 0;
}
