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;
bool mark[70];
int main() {
  long long u, v;
  cin >> u >> v;
  if (u > v) return cout << "-1", 0;
  if (u == v) {
    if (u) return cout << "1" << endl << u << endl, 0;
    return cout << "0", 0;
  }
  vector<int> a, b;
  for (long long i = 0; i < 63; i++)
    if (u & (1LL << i)) a.push_back(i), mark[i] = 1;
  long long diff = v - u;
  if (diff & 1) return cout << "-1", 0;
  diff /= 2;
  vector<int> lagbe;
  int cnt = 2;
  for (int i = 0; i < 63; i++)
    if (diff & (1LL << i)) {
      lagbe.push_back(i);
      if (mark[i]) cnt = min(cnt + 1, 3);
    }
  long long ansx = 0, ansy = 0, ansz = 0;
  for (int i = 0; i < a.size(); i++) ansx |= 1LL << a[i];
  for (auto x : lagbe) ansy |= 1LL << x;
  if (cnt > 2)
    for (auto x : lagbe) ansz |= 1LL << x;
  else
    for (auto x : lagbe) ansx |= 1LL << x;
  cout << cnt << endl;
  cout << ansx << " " << ansy;
  if (cnt > 2) cout << " " << ansz;
}
