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;
int p[80];
unsigned long long o[80];
int main() {
  int i, j;
  unsigned long long n, m, k, a, b, x, kk;
  scanf("%lld%lld", &n, &m);
  if (n == m) {
    if (n == 0)
      printf("0\n");
    else
      printf("1\n%lld\n", n);
    return 0;
  }
  if (n > m) {
    printf("-1\n");
    return 0;
  }
  k = n;
  i = 1;
  x = 1;
  kk = m - n;
  while (k || x < kk) {
    if (k & 1) {
      p[i] = 1;
    } else
      p[i] = 0;
    k /= 2;
    o[i] = x;
    x *= 2;
    i++;
  }
  o[i] = x;
  int max = i;
  unsigned long long ans = 0;
  for (i = max; i > 1; i--) {
    if (kk >= o[i]) {
      p[i - 1] += 2;
      kk -= o[i];
      if (kk == 0) break;
    }
  }
  if (kk) {
    printf("-1\n");
    return 0;
  }
  int xum = 0;
  for (i = 1; i <= max; i++) {
    if (p[i]) xum = i;
    if (p[i] > ans) ans = p[i];
  }
  printf("%lld\n", ans);
  for (i = 1; i <= ans; i++) {
    unsigned long long q = 0;
    for (j = xum; j >= 1; j--) {
      if (p[j]) {
        q += o[j];
        p[j]--;
      }
    }
    if (i == 1)
      printf("%lld", q);
    else
      printf(" %lld", q);
  }
  return 0;
}
