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;
const long long inf = 1e17 + 9;
const long long mod = 1e9 + 7;
const long long maxn = 2e5 + 8;
long long u, v;
void solve() {
  long long i, j;
  scanf("%lld%lld", &u, &v);
  if (u > v || (u + v) & 1) {
    printf("-1\n");
    return;
  }
  if (u == v) {
    if (u == 0)
      printf("0\n");
    else
      printf("1\n%lld\n", u);
    return;
  }
  long long x = (v - u) / 2;
  if (u & x)
    printf("3\n%lld %lld %lld\n", u, x, x);
  else
    printf("2\n%lld %lld\n", (u + x), x);
}
signed main() {
  long long t = 1;
  while (t--) solve();
  return 0;
}