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 fast() {
  cin.tie(0);
  cin.sync_with_stdio(0);
}
vector<string> v;
bool check(string s) {
  int freq[26] = {0};
  for (int i = 0; i < s.length(); ++i) {
    freq[s[i] - 'A']++;
    if (freq[s[i] - 'A'] > 1) return false;
  }
  return true;
}
long long fastExpo(long long a, long long b) {
  if (!b) return 1;
  long long half = fastExpo(a, b / 2);
  long long ans = (half * half);
  if (b & 1) return (ans * a);
  return ans;
}
int main() {
  fast();
  long long a, b;
  cin >> a >> b;
  if (a == b) {
    if (a == 0)
      cout << a;
    else
      cout << 1 << "\n" << a;
  } else if (a > b || (b - a) % 2 != 0)
    cout << -1;
  else {
    long long temp = (b - a) / 2;
    if ((((temp + a) ^ temp) == a))
      cout << 2 << "\n" << temp + a << " " << temp;
    else
      cout << 3 << "\n" << a << " " << temp << " " << temp;
  }
}
