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 read() {
  char ch = getchar();
  long long f = 1, x = 0;
  while (ch < '0' || ch > '9') {
    f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return f * x;
}
long long n;
signed main() {
  n = 1;
  for (long long i = 1; i <= n; i++) {
    long long u = read();
    long long v = read();
    if (v < u) {
      cout << -1 << endl;
    } else if (u == 0 && v == 0) {
      cout << 0 << endl;
    } else if (u == v) {
      cout << 1 << endl;
      cout << u << endl;
    } else if (u % 2 == v % 2) {
      long long p = (v - u) / 2, q = (u + v) / 2;
      if ((p ^ q) == u) {
        cout << 2 << endl;
        cout << ((v - u) / 2) << " " << (u + v) / 2 << endl;
      } else {
        cout << 3 << endl;
        cout << u << " " << (v - u) / 2 << " " << (v - u) / 2 << endl;
      }
    } else {
      cout << -1 << endl;
    }
  }
  return 0;
}
