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 i, u, v, x, flag, n1, n2;
  cin >> u >> v;
  if (u > v) {
    cout << -1 << endl;
    return 0;
  }
  if (u == v) {
    if (u == 0) {
      cout << 0 << endl;
      return 0;
    }
    cout << 1 << endl;
    cout << u << endl;
    return 0;
  }
  if ((v - u) % 2 != 0) {
    cout << -1 << endl;
    return 0;
  }
  vector<long long> a, b;
  x = u;
  while (x > 0) {
    a.push_back(x % 2);
    x /= 2;
  }
  x = v - u;
  while (x > 0) {
    b.push_back(x % 2);
    x /= 2;
  }
  n1 = a.size();
  n2 = b.size();
  if (n1 < n2) {
    for (i = 1; i <= n2 - n1; i += 1) a.push_back(0);
  }
  flag = 1;
  for (i = 1; i < n2; i += 1) {
    if (b[i] == 1 && a[i - 1] != 0) {
      flag = 0;
      break;
    }
  }
  x = v - u;
  if (flag == 1) {
    cout << 2 << endl;
    cout << u + x / 2 << " " << x / 2 << endl;
  } else {
    cout << 3 << endl;
    cout << u << " " << x / 2 << " " << x / 2 << endl;
  }
}
