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;
void __print(int x) { cerr << x; }
void __print(long x) { cerr << x; }
void __print(long long x) { cerr << x; }
void __print(unsigned x) { cerr << x; }
void __print(unsigned long x) { cerr << x; }
void __print(unsigned long long x) { cerr << x; }
void __print(float x) { cerr << x; }
void __print(double x) { cerr << x; }
void __print(long double x) { cerr << x; }
void __print(char x) { cerr << '\'' << x << '\''; }
void __print(const char *x) { cerr << '\"' << x << '\"'; }
void __print(const string &x) { cerr << '\"' << x << '\"'; }
void __print(bool x) { cerr << (x ? "true" : "false"); }
template <typename T, typename V>
void __print(const pair<T, V> &x) {
  cerr << '{';
  __print(x.first);
  cerr << ',';
  __print(x.second);
  cerr << '}';
}
template <typename T>
void __print(const T &x) {
  int first = 0;
  cerr << '{';
  for (auto &i : x) cerr << (first++ ? "," : ""), __print(i);
  cerr << "}";
}
void _print() { cerr << "]\n"; }
template <typename T, typename... V>
void _print(T t, V... v) {
  __print(t);
  if (sizeof...(v)) cerr << ", ";
  _print(v...);
}
const long long INF = 1e18;
long long mod = 1000000007;
inline long long fast_expo(long long base, long long power,
                           long long modulo = mod) {
  base %= modulo;
  if (base < 0) base += modulo;
  long long x = base, cnt = power, ans = 1;
  while (cnt) {
    if (cnt & 1) ans = (ans * x) % modulo;
    x = (x * x) % modulo;
    cnt >>= 1;
  }
  return ans;
}
long long takemod(long long a, long long mo) {
  a %= mo;
  if (a < 0) a += mo;
  return a;
}
long long modinv(long long a, long long mo) {
  return takemod(fast_expo(takemod(a, mo), mo - 2, mo), mo);
}
int main() {
  long long u, v;
  cin >> u >> v;
  if (u > v || ((u & 1) != (v & 1))) {
    cout << -1 << "\n";
    return 0;
  }
  if (u == 0 && v == 0) {
    cout << 0 << endl;
    return 0;
  }
  if (u == v) {
    cout << 1 << "\n" << u << "\n";
    return 0;
  }
  long long a = (v - u) / 2;
  long long b = a + u;
  if ((b ^ a) == u) {
    cout << 2 << endl;
    cout << a << " " << b << endl;
    return 0;
  } else {
    cout << 3 << endl;
    cout << a << " " << a << " " << u << endl;
  }
  return 0;
}
