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;
template <class T>
int chkmax(T& a, T b) {
  if (b > a) {
    a = b;
    return 1;
  }
  return 0;
}
template <class T>
int chkmin(T& a, T b) {
  if (b < a) {
    a = b;
    return 1;
  }
  return 0;
}
template <class iterator>
void output(iterator begin, iterator end, ostream& out = cerr) {
  while (begin != end) {
    out << (*begin) << " ";
    begin++;
  }
  out << endl;
}
template <class T>
void output(T x, ostream& out = cerr) {
  output(x.begin(), x.end(), out);
}
void fast_io() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
}
long long xr, sum;
signed main() {
  fast_io();
  cin >> xr >> sum;
  if (sum < xr || sum % 2 != xr % 2) {
    cout << -1 << endl;
    exit(0);
  }
  if (sum == 0 && xr == 0) {
    cout << 0 << endl;
    exit(0);
  }
  if (sum == xr) {
    cout << 1 << endl;
    cout << sum << endl;
    exit(0);
  }
  long long _sum = sum;
  long long a = xr, b = 0;
  for (int i = 61; i >= 0; --i) {
    if (((xr >> i) & 1LL) == 0) {
      long long val = (1LL << i);
      if (sum - xr >= 2 * val) {
        sum -= 2 * val;
        a += val;
        b += val;
      }
    }
  }
  if (sum == xr) {
    cout << 2 << endl;
    cout << a << " " << b << endl;
    exit(0);
  }
  cout << 3 << endl;
  cout << xr << " " << (_sum - xr) / 2 << " " << (_sum - xr) / 2 << endl;
}
