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;
vector<long long> ans;
long long cnt[64];
signed main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  long long n, m, i, j;
  cin >> n >> m;
  if (((m - n) & 1) || n > m) {
    cout << "-1\n";
    return 0;
  }
  for (i = 0; i < 63; i++) {
    if ((n >> i) & 1) {
      cnt[i]++;
    }
  }
  for (i = 1; i < 63; i++) {
    if (((m - n) >> i) & 1) {
      cnt[i - 1] += 2;
    }
  }
  for (i = 0; i < 63; i++) {
    for (j = 0; j < cnt[i]; j++) {
      if (j == ans.size()) {
        ans.emplace_back(0);
      }
      ans[j] += (1LL << i);
    }
  }
  cout << ans.size() << '\n';
  for (long long aoi : ans) {
    cout << aoi << ' ';
  }
  cout << '\n';
  return 0;
}
