Problem: 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.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int main() {
  long long n, m, x = 0, y = 0;
  scanf("%lld %lld", &n, &m);
  bool imp = false;
  for (long long i = 0, j = 1; i < 63; i++, j <<= 1) {
    if ((n & j) != 0) x |= j;
    if (((n & j) != 0) != ((m & j) != 0)) {
      if (i == 0) {
        imp = true;
        break;
      }
      y |= j >> 1;
      if ((n & j) != 0) m -= (j << 1);
      if (m < 0) {
        imp = true;
        break;
      }
    }
  }
  if (imp) {
    printf("-1\n");
  } else if (x == 0) {
    if (y == 0)
      printf("0\n");
    else
      printf("2\n%lld %lld\n", y, y);
  } else if (y == 0) {
    printf("1\n%lld\n", x);
  } else if ((x ^ y) == x + y) {
    printf("2\n%lld %lld\n", x + y, y);
  } else {
    printf("3\n%lld %lld %lld\n", x, y, y);
  }
}