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;
long long int mypow(long long int a, long long int b) {
  long long int res = 1LL;
  while (b) {
    if (b & 1) res = res * a % 1000000007;
    a = a * a % 1000000007;
    b >>= 1;
  }
  return res;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int t = 1;
  while (t--) {
    long long int u, v;
    cin >> u >> v;
    if (u > v || (v - u) % 2 == 1) {
      cout << -1 << endl;
      continue;
    }
    if (u == v && u == 0) {
      cout << 0 << endl;
      continue;
    }
    if (u == v) {
      cout << 1 << endl;
      cout << u << endl;
      continue;
    }
    long long int x = (v - u) / 2;
    if ((u & x) == 0) {
      cout << 2 << endl;
      cout << (u + x) << " " << x << endl;
    } else {
      cout << 3 << endl;
      cout << u << " " << x << " " << x << endl;
    }
  }
  return 0;
}