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;
long long n, ans[100001], curr, deg[100001] = {0};
vector<pair<int, int>> adj[100001];
void dfs(long long u, long long p) {
  for (auto v : adj[u]) {
    if (v.first == p) {
      continue;
    }
    if (ans[v.second] == -1) {
      ans[v.second] = curr++;
    }
    dfs(v.first, u);
  }
}
void solve() {
  long long n;
  cin >> n;
  long long m;
  cin >> m;
  if (n > m || n % 2 != m % 2) {
    cout << -1 << endl;
    return;
  }
  if (n == m && n == 0) {
    cout << n << endl;
    return;
  }
  if (n == m) {
    cout << 1 << endl << n << endl;
    return;
  }
  long long x = (m - n) / 2;
  if ((x & n) == 0) {
    cout << 2 << endl << n + x << " " << x << endl;
    return;
  }
  {
    cout << 3 << endl << n << " " << x << " " << x << endl;
    return;
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  solve();
  return 0;
}
