Problem: 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.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
map<pair<unsigned long long, unsigned long long>, unsigned long long> dp;
unsigned long long lazy_dp(unsigned long long u, unsigned long long v) {
  auto key = make_pair(u, v);
  if (dp.count(key)) return dp[key];
  unsigned long long ans = -1;
  if (u == v)
    ans = u;
  else if ((u & 1) != (v & 1))
    ans = -1;
  else {
    if (u & 1) {
      unsigned long long r = lazy_dp(u / 2, (v - 1) / 2);
      if (r != -1) ans = (r * 2);
    } else {
      unsigned long long r1 = lazy_dp(u / 2, (v - 2) / 2);
      unsigned long long r2 = lazy_dp(u / 2, v / 2);
      if (r1 != -1)
        ans = 1 + (r1 * 2);
      else if (r2 != -1)
        ans = r2 * 2;
    }
  }
  dp[key] = ans;
  return ans;
}
int main(int argc, char **argv) {
  ios::sync_with_stdio(false);
  cin.tie(0);
  unsigned long long u, v;
  cin >> u >> v;
  if (u > v || (u & 1) != (v & 1)) {
    cout << "-1\n";
    return 0;
  }
  if (u == v) {
    if (u == 0)
      cout << "0\n";
    else
      cout << "1\n" << u << "\n";
    return 0;
  }
  auto p = lazy_dp(u, v);
  if (p != -1)
    cout << "2\n" << p << " " << (u ^ p) << "\n";
  else {
    unsigned long long diff = (v - u) / 2;
    cout << "3\n" << u << " " << diff << " " << diff << "\n";
  }
  return 0;
}