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>
const long long mod = (long long)1e9 + 7;
using namespace std;
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long u, v;
  cin >> u >> v;
  if (u == 0 and v == 0) {
    cout << 0 << "\n";
    return 0;
  }
  if (u == v) {
    cout << 1 << "\n";
    cout << u << "\n";
    return 0;
  }
  if (u > v or (v - u) % 2) {
    cout << -1 << "\n";
    return 0;
  }
  vector<long long> bits(61, 0);
  for (long long i = 61; i >= 0; i--) {
    if (u & (1ll << i)) {
      bits[i] = 1;
    }
  }
  long long diff = v - u;
  vector<long long> bb;
  for (long long i = 1; i <= (long long)1e18; i *= 2) bb.push_back(i);
  long long i1, i2, i3;
  i1 = i2 = i3 = 0;
  i1 = u;
  while (diff) {
    long long idx = upper_bound(bb.begin(), bb.end(), diff) - bb.begin();
    idx--;
    long long val = bb[idx];
    if (bits[log2(val / 2)] == 1) {
      i2 = i2 | (val / 2);
      i3 = i3 | (val / 2);
    } else {
      i1 = i1 | (val / 2);
      i2 = i2 | (val / 2);
    }
    diff -= val;
  }
  if (i3 == 0) {
    cout << 2 << "\n";
    cout << i1 << ' ' << i2 << "\n";
  } else {
    cout << 3 << "\n";
    cout << i1 << ' ' << i2 << ' ' << i3 << "\n";
  }
  return 0;
}
