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() {
  ios_base::sync_with_stdio(0);
  long long u, v;
  cin >> u >> v;
  vector<long long> rtn;
  if (u > v) {
    cout << -1 << "\n";
    return 0;
  }
  if (u == 0 && v == 0) {
    cout << 0 << "\n";
    return 0;
  }
  rtn.push_back(u);
  long long dif = v - u;
  if (1 == dif % 2) {
    cout << -1 << "\n";
    return 0;
  }
  long long val = 0;
  for (long long i = 0; i <= 63; i++) {
    if (dif & (1ll << i)) {
      val += (1ll << (i - 1));
    }
  }
  rtn.push_back(val);
  bool add = true;
  for (long long i = 0; i <= 63; i++) {
    if (val & (1ll << i) && rtn[0] & (1ll << i)) {
      add = false;
    }
  }
  if (add) {
    rtn[0] += val;
  } else {
    rtn.push_back(val);
  }
  long long sz = 0;
  for (int i = 0; i < rtn.size(); i++) {
    if (rtn[i] != 0) {
      sz++;
    }
  }
  cout << sz << "\n";
  for (int i = 0; i < rtn.size(); i++) {
    if (rtn[i] != 0) {
      cout << rtn[i] << " ";
    }
  }
  return 0;
}
