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;
void Fast() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
}
void File() {}
long long gcd(long long a, long long b) { return !b ? a : gcd(b, a % b); }
int main() {
  Fast(), File();
  long long u, v;
  cin >> u >> v;
  if (v < u) {
    cout << -1 << "\n";
    return 0;
  }
  vector<long long> solve;
  solve.push_back(u);
  v -= u;
  if (v & 1) solve.push_back(1);
  if (v / 2) solve.push_back(v / 2), (v / 2), solve.push_back(v / 2);
  int n = solve.size();
  long long sum = 0, xorres = 0;
  int mn = INT_MAX;
  vector<long long> res, finalres;
  for (int i = 0; i < (1 << n); i++) {
    sum = 0, res.clear(), xorres = 0;
    for (int j = 0; j < n; j++) {
      if (i & (1 << j)) {
        sum += solve[j];
      } else
        res.push_back(solve[j]);
    }
    if (sum) res.push_back(sum);
    for (int idx = 0; idx < res.size(); idx++) {
      xorres ^= res[idx];
    }
    if (xorres == u) {
      if (res.size() < mn) {
        mn = res.size();
        finalres = res;
      }
    }
  }
  if (mn == INT_MAX)
    cout << -1 << "\n";
  else {
    cout << mn << "\n";
    for (int i = 0; i < finalres.size(); i++) cout << finalres[i] << " ";
  }
}
