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;
const long long MODULO = 1000000007;
void ftime() {}
static int speed_increase = []() {
  std::ios::sync_with_stdio(false);
  std::cin.tie(NULL);
  return 0;
}();
vector<long long> arr;
void solve() {
  long long u, v;
  cin >> u >> v;
  if (u > v) {
    cout << -1;
    return;
  }
  if (u == v) {
    if (v == 0) {
      cout << "0";
    } else {
      cout << "1\n" << v;
    }
    return;
  }
  long long x = v - u;
  if (x % 2) {
    cout << -1;
    return;
  }
  x >>= 1;
  while ((x & u) != 0) {
    x &= u;
    arr.push_back(x);
    v -= x;
    u ^= x;
    x = v - u;
    if (x % 2) {
      cout << -1;
      return;
    }
    x >>= 1;
  }
  long long y = x | u;
  arr.push_back(x);
  arr.push_back(y);
  cout << arr.size() << "\n";
  for (auto &val : arr) {
    cout << val << " ";
  }
}
int main() {
  solve();
  return 0;
}
