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;
void test_case() {
  long long int u, v;
  cin >> u >> v;
  if (v < u) {
    cout << -1 << endl;
    return;
  }
  long long int tp = v - u, t = u;
  long long int res[62] = {};
  for (long long int i = 0; i < 62; i++) {
    res[i] += t % 2;
    t /= 2;
  }
  long long int ar[62];
  ar[0] = 1;
  for (long long int i = 1; i < 62; i++) ar[i] = 2 * ar[i - 1];
  while (tp > 0) {
    if (tp == 1) {
      cout << -1 << endl;
      return;
    }
    for (long long int i = 61; i >= 0; i--) {
      long long int x = ar[i];
      if (tp / x >= 2) {
        res[i] += 2;
        tp -= 2 * x;
      }
    }
  }
  long long int mx = -1;
  for (long long int i = 0; i < 62; i++) {
    mx = max(mx, res[i]);
  }
  cout << mx << endl;
  for (long long int j = 0; j < mx; j++) {
    long long int nm = 0;
    for (long long int i = 0; i < 62; i++) {
      if (res[i] > 0) {
        nm += ar[i];
        res[i]--;
      }
    }
    cout << nm << " ";
  }
  cout << endl;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int t = 1;
  while (t--) {
    test_case();
  }
  return 0;
}
