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 const INF = 1000000007;
long long mod = 1e9 + 7;
long long gcd(long long a, long long b) {
  if (a == 0) return b;
  if (b == 0) return a;
  return gcd(b, a % b);
}
struct hash_pair {
  template <class T1, class T2>
  size_t operator()(const pair<T1, T2>& p) const {
    auto hash1 = hash<T1>{}(p.first);
    auto hash2 = hash<T2>{}(p.second);
    return hash1 ^ hash2;
  }
};
struct custom_hash {
  static uint64_t splitmix64(uint64_t x) {
    x += 0x9e3779b97f4a7c15;
    x = (x ^ (x >> 30)) * 0xbf58476d1ce4e5b9;
    x = (x ^ (x >> 27)) * 0x94d049bb133111eb;
    return x ^ (x >> 31);
  }
  size_t operator()(uint64_t x) const {
    static const uint64_t FIXED_RANDOM =
        chrono::steady_clock::now().time_since_epoch().count();
    return splitmix64(x + FIXED_RANDOM);
  }
};
bool isPerfectSquare(long double x) {
  if (x >= 0) {
    long long sr = sqrt(x);
    return (sr * sr == x);
  }
  return false;
}
long long fast_pow(long long a, long long p) {
  long long res = 1;
  while (p) {
    if (p % 2 == 0) {
      a = (a * a) % mod;
      p /= 2;
    } else {
      res = (res * a) % mod;
      p--;
    }
  }
  return res;
}
void test() {
  long long a, b;
  cin >> a >> b;
  if (a > b or (b - a) % 2) {
    cout << -1 << "\n";
    return;
  }
  if (b == 0) {
    cout << 0 << "\n";
    return;
  }
  if (a == b) {
    cout << 1 << "\n";
    cout << a << "\n";
    return;
  }
  long long diff = (b - a) / 2;
  if ((diff ^ a) == (diff + a)) {
    cout << 2 << "\n";
    cout << diff << " " << (diff ^ a) << "\n";
    return;
  }
  cout << 3 << "\n";
  cout << a << " " << diff << " " << diff << "\n";
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long t = 1;
  long long i = 1;
  while (t--) {
    test();
    i++;
  }
  return 0;
}
