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>
#pragma GCC optimize("O3")
using namespace std;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long int u, v, a, b, c, d, e;
  cin >> u >> v;
  if (v < u || (v - u) % 2 == 1) {
    cout << -1;
    return 0;
  }
  a = u;
  b = v - u;
  long long int arr[70];
  long long int i = 0;
  memset(arr, 0, sizeof(arr));
  while (a) {
    if (a & 1) {
      arr[i]++;
    }
    i++;
    a = a / 2;
  }
  i = 0;
  while (b) {
    if (b & 1) {
      arr[i - 1] += 2;
    }
    i++;
    b = b / 2;
  }
  long long int ans = 0;
  for (long long int i = 0; i < 70; i++) {
    ans = max(ans, arr[i]);
  }
  e = 0;
  cout << ans << '\n';
  for (long long int i = 0; i < ans; i++) {
    e = 0;
    for (long long int j = 0; j < 70; j++) {
      if (arr[j] > 0) {
        e = e + pow(2, j);
        arr[j]--;
      }
    }
    cout << e << " ";
  }
  return 0;
}
