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 u, v;
long long a[10];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int tCase = 1;
  while (tCase--) {
    cin >> u >> v;
    if (u == 0 && v == 0) {
      cout << 0;
      continue;
    }
    if (u > v) {
      cout << -1 << endl;
      continue;
    }
    a[1] = u;
    v -= u;
    if (v & 1) {
      cout << -1 << endl;
      continue;
    }
    long long cur = 1;
    while (v > 0) {
      long long nxt = cur * 2;
      if (v & nxt) {
        for (int i = 1; i <= 2; i++)
          for (int j = 1; j <= 3; j++)
            if ((a[j] & cur) == 0) {
              a[j] += cur;
              break;
            }
        v -= nxt;
      }
      cur = nxt;
    }
    int ans = 1;
    while (a[ans + 1] > 0) ans++;
    cout << ans << endl;
    for (int i = 1; i <= ans; i++) cout << a[i] << ' ';
  }
  return 0;
}
