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;
int main() {
  long long u, v, x;
  cin >> u >> v;
  bool pos = 1;
  if (v < u || v % 2 != u % 2) pos = 0;
  v -= u;
  x = 0;
  for (long long i = 1; i < 63 && pos; i++) {
    bool bu = u & (1ll << i), bv = v & (1ll << i), bx = x & (1ll << i);
    bool bun = u & (1ll << (i - 1)), bvn = v & (1ll << (i - 1)),
         bxn = x & (1ll << (i - 1));
    if (bv == 1) {
      v ^= (1ll << i);
      if (bun == 0 && bvn == 0) {
        u |= (1ll << (i - 1));
        v |= (1ll << (i - 1));
      } else if (bun == 1 && bvn == 0) {
        v |= (1ll << (i - 1));
        x |= (1ll << (i - 1));
      }
    }
  }
  if (!pos) {
    cout << -1 << endl;
  } else if (u == 0 && v == 0) {
    cout << 0 << endl;
  } else if (v == 0) {
    cout << 1 << endl;
    cout << u << endl;
  } else if (x == 0) {
    cout << 2 << endl;
    cout << u << " " << v << endl;
  } else {
    cout << 3 << endl;
    cout << u << " " << v << " " << x << endl;
  }
  return 0;
}
