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;
using LL = long long;
int main() {
  LL x, s;
  cin >> x >> s;
  LL d = s - x;
  if (d >= 0 && d % 2 == 0) {
    int K = 60;
    vector<int> c(K, 0);
    for (int j = 1; j < K; ++j) {
      if (d & ((LL)1 << j)) {
        c[j - 1] += 2;
      }
    }
    for (int j = 0; j < K; ++j) {
      if (x & ((LL)1 << j)) {
        c[j] += 1;
      }
    }
    vector<LL> ans;
    while (true) {
      LL y = 0;
      for (int j = 0; j < K; ++j) {
        if (c[j] > 0) {
          y |= ((LL)1 << j);
          --c[j];
        }
      }
      if (y == 0) {
        break;
      } else {
        ans.push_back(y);
      }
    }
    cout << ans.size() << '\n';
    for (int j = 0; j < ans.size(); ++j) {
      cout << ans[j] << ((j == ans.size() - 1) ? '\n' : ' ');
    }
  } else {
    cout << "-1\n";
  }
  return 0;
}
