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 xr, sm, df, sq[3], zs;
int main() {
  long long i;
  scanf("%lld%lld", &xr, &sm);
  df = sm - xr;
  if (df < 0 || df % 2) {
    zs = -1;
  } else if (!df) {
    if (!xr) {
      zs = 0;
    } else {
      zs = 1;
      sq[0] = xr;
    }
  } else {
    df /= 2;
    if ((xr ^ df) == (xr | df)) {
      zs = 2;
      sq[0] = xr | df;
      sq[1] = df;
    } else {
      zs = 3;
      sq[0] = xr;
      sq[1] = df;
      sq[2] = df;
    }
  }
  printf("%lld\n", zs);
  for (i = 0; i < zs; i++) {
    printf("%lld%c", sq[i], " \n"[i == zs - 1]);
  }
}
