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 jc[100];
int main() {
  long long i, j, n, m;
  jc[1] = 2;
  long long bz = 1000000000000000000ll;
  int jz;
  cin >> n >> m;
  for (i = 2; 1; i++) {
    jc[i] = jc[i - 1] * 2;
    if (jc[i] > bz) {
      jz = i;
      break;
    }
  }
  if (n == 0 && m == 0)
    cout << 0 << endl;
  else if (n > m || (m - n) % 2 == 1) {
    cout << -1 << endl;
  } else if (n == m) {
    cout << 1 << endl;
    cout << n << endl;
  } else {
    int cnt = 1;
    long long jp = 0;
    long long dx = m - n;
    jp = dx / 2;
    if (((n + jp) ^ jp) == n) {
      printf("2\n");
      printf("%lld %lld\n", n + jp, jp);
    } else {
      printf("3\n");
      printf("%lld %lld %lld\n", n, jp, jp);
    }
  }
  return 0;
}
