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;
const int maxn = (int)2e5 + 100;
const double eps = 1e-7;
int lowbit(int x) { return x & (-x); }
struct P {
  int x, y;
};
vector<long long> bit;
int main() {
  long long u, v;
  scanf("%lld %lld", &u, &v);
  if (u == 0 && v == 0) {
    cout << 0 << endl;
    return 0;
  }
  bit.push_back(u);
  long long sub = v - u;
  if (sub < 0 || sub % 2 == 1)
    cout << "-1" << endl;
  else {
    sub >>= 1;
    for (int i = 62; i >= 0; i--) {
      long long a = 1LL << i;
      if (a & sub) {
        int flag = 2;
        for (int i = 0; i < bit.size(); i++) {
          if ((bit[i] & a) == 0) {
            flag--;
            bit[i] += a;
          }
          if (flag == 0) break;
        }
        while (flag--) bit.push_back(a);
      }
    }
    cout << bit.size() << endl;
    for (auto i : bit) cout << i << " ";
  }
}
