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 err(istream_iterator<string> it) { cout << ",,,,,,,,,,,,,\n"; }
template <typename T, typename... Args>
void err(istream_iterator<string> it, T a, Args... args) {
  cerr << *it << " = " << a << ", ";
  err(++it, args...);
}
template <typename... T>
void rd(T&... args) {
  ((cin >> args), ...);
}
template <typename... T>
void ps(T... args) {
  ((cout << args << ' '), ...);
  cout << '\n';
}
const int MOD = 1e9 + 7, MOD1 = 998244353LL, MAX = 1e5 + 5;
const char nl = '\n';
const int INF = 1e9 + 5;
long long u, v;
void solve() {
  cin >> u >> v;
  if (u > v || (u % 2 != v % 2)) {
    cout << "-1\n";
    return;
  }
  if (v == 0) {
    cout << "0\n";
    return;
  }
  if (u == v) {
    cout << "1\n" << u << '\n';
    return;
  }
  long long x = (v - u) / 2, var = (x ^ u);
  if (var + x == v) {
    cout << "2\n" << x << ' ' << var << '\n';
    return;
  }
  cout << "3\n" << x << ' ' << x << ' ' << u << '\n';
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  solve();
  return 0;
}
