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 powmod(long long base, long long exponent, long long mod) {
  long long ans = 1;
  while (exponent) {
    if (exponent & 1) ans = (ans * base) % mod;
    base = (base * base) % mod;
    exponent /= 2;
  }
  return ans;
}
long long gcd(long long a, long long b) {
  if (b == 0)
    return a;
  else
    return gcd(b, a % b);
}
const int upperlimit = 1e6 + 100;
const int mod = 1e9 + 7;
int ans[64];
long long gg[3];
int s[64];
int x[64];
int carry[64];
int main() {
  long long sum, xr;
  scanf("%lld", &xr);
  scanf("%lld", &sum);
  if ((xr > sum) || ((sum & 1) != (xr & 1))) {
    printf("%d", -1);
    return 0;
  }
  if (sum == xr) {
    if (sum == 0) {
      printf("%d", 0);
      return 0;
    }
    printf("%d\n", 1);
    printf("%lld", sum);
    return 0;
  }
  long long o = sum + xr;
  o >>= 1;
  long long a = sum - xr;
  a >>= 1;
  if ((o & a) == a) {
    printf("%d\n", 2);
    printf("%lld ", o);
    printf("%lld", a);
    return 0;
  }
  printf("%d\n", 3);
  sum -= xr;
  sum >>= 1;
  printf("%lld ", xr);
  printf("%lld ", sum);
  printf("%lld", sum);
  return 0;
}
