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;
const int N = 100005;
const int inf = 1000 * 1000 * 1000;
const int mod = 1000 * 1000 * 1000 + 7;
mt19937 myrand(chrono::steady_clock::now().time_since_epoch().count());
long long a, b;
int main() {
  cin >> a >> b;
  if (a == b && a == 0) {
    cout << 0 << endl;
    return 0;
  }
  if (a == b) {
    cout << 1 << endl;
    cout << a << endl;
    return 0;
  }
  if (a == 0) {
    if (b % 2) {
      cout << -1 << endl;
      return 0;
    } else {
      cout << 2 << endl;
      cout << b / 2 << " " << b / 2 << endl;
    }
    return 0;
  }
  if (b == 0) {
    cout << -1 << endl;
    return 0;
  }
  if (a > b || a % 2 != b % 2) {
    cout << -1 << endl;
    return 0;
  }
  long long x = (b - a) / 2;
  if ((a + x) == (a | x)) {
    cout << 2 << endl;
    cout << x << " " << a + x << endl;
  } else {
    cout << 3 << endl;
    cout << a << " " << x << " " << x << endl;
  }
  return 0;
}