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);
  cin.tie(0);
  cout.tie(0);
  int64_t x, sum;
  cin >> x >> sum;
  int64_t xp = (x & (int64_t)1);
  int64_t sp = (sum & (int64_t)1);
  if (xp != sp) {
    cout << -1 << endl;
    return 0;
  }
  vector<bool> arr[64];
  int64_t carry = 0;
  for (int64_t bit = 0; bit < 64; ++bit) {
    int64_t xb = ((x >> bit) & (int64_t)1);
    int64_t sb = ((sum >> bit) & (int64_t)1);
    if (carry == 0 && xb == sb) {
      if (xb) {
        arr[bit].push_back(1);
      }
    } else if (carry == 0 && xb != sb) {
      arr[bit - 1].push_back(1);
      arr[bit - 1].push_back(1);
      if (xb) {
        arr[bit].push_back(1);
        carry = 1;
      }
    } else if (carry == 1 && xb == sb) {
      arr[bit - 1].push_back(1);
      arr[bit - 1].push_back(1);
      if (xb) {
        arr[bit].push_back(1);
      }
    } else if (carry == 1 && xb != sb) {
      if (xb) {
        arr[bit].push_back(1);
        carry = 1;
      } else {
        carry = 0;
      }
    }
  }
  if (carry == 1) {
    cout << -1 << endl;
    return 0;
  }
  int64_t n = 0;
  for (int64_t i = 0; i < 64; i++) {
    n = max(n, (int64_t)arr[i].size());
  }
  vector<int64_t> ans(n, 0);
  for (int64_t bit = 0; bit < 64; ++bit) {
    int64_t mask = 1;
    mask = (mask << bit);
    for (int64_t i = 0; i < arr[bit].size(); i++) {
      ans[i] = (ans[i] | mask);
    }
  }
  cout << n << endl;
  for (int64_t i = 0; i < n; i++) {
    cout << ans[i] << " ";
  }
  cout << endl;
  return 0;
}
