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;
int main() {
  long long int u, v, v0;
  cin >> u >> v;
  v0 = v;
  long long int x[70] = {0}, help[70] = {0};
  int i = 0;
  while (u != 0) {
    x[i] = u % 2;
    help[i] = x[i];
    if (x[i]) v -= ((long long int)1 << i);
    if (v < 0) break;
    u /= 2;
    i++;
  }
  if (v < 0 or v % 2 != 0)
    cout << -1;
  else {
    i = 0;
    v /= 2;
    while (v != 0) {
      help[i] += 2 * (v % 2);
      v /= 2;
      i++;
    }
    long long int maxx = 0;
    for (i = 0; i <= 60; i++) maxx = max(help[i], maxx);
    cout << maxx << endl;
    for (i = 1; i <= maxx; i++) {
      long long int ans = 0;
      for (int j = 0; j <= 60; j++) {
        if (help[j] >= i) ans = ans + ((long long int)1 << j);
      }
      cout << ans << " ";
    }
  }
}
