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;
bool comp(pair<int, int> a, pair<int, int> b) {
  return (a.first == b.first) ? a.second < b.second : a.first < b.first;
}
long long int modInverse(long long int a) {
  long long int m = 1000000007;
  long long int y = 0, x = 1;
  while (a > 1) {
    long long int q = a / m;
    long long int t = m;
    m = a % m, a = t;
    t = y;
    y = x - q * y;
    x = t;
  }
  if (x < 0) x += 1000000007;
  return x;
}
void solve() {
  long long int u, v;
  cin >> u >> v;
  if (u > v) {
    cout << -1 << '\n';
  } else if (u + v == 0) {
    cout << 0 << '\n';
  } else if (u == v) {
    cout << 1 << '\n' << u << '\n';
  } else {
    if (!u) {
      if (v % 2)
        cout << -1 << '\n';
      else
        cout << 2 << '\n' << v / 2 << " " << v / 2 << '\n';
    } else {
      long long int out = (v - u) / 2;
      if ((v - u) % 2) {
        cout << -1 << '\n';
      } else {
        if (!(u & out)) {
          cout << 2 << '\n' << u + out << " " << out << '\n';
        } else
          cout << 3 << '\n' << u << " " << out << " " << out << '\n';
      }
    }
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  int t = 1, nw = 1;
  while (t--) {
    solve();
  }
}
