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) { cout << x; }
void __print(long x) { cout << x; }
void __print(long long x) { cout << x; }
void __print(unsigned x) { cout << x; }
void __print(unsigned long x) { cout << x; }
void __print(unsigned long long x) { cout << x; }
void __print(float x) { cout << x; }
void __print(double x) { cout << x; }
void __print(long double x) { cout << x; }
void __print(char x) { cout << '\'' << x << '\''; }
void __print(const char *x) { cout << '\"' << x << '\"'; }
void __print(const string &x) { cout << '\"' << x << '\"'; }
void __print(bool x) { cout << (x ? "true" : "false"); }
template <typename T, typename V>
void __print(const pair<T, V> &x) {
  __print(x.first);
  cout << ',';
  __print(x.second);
}
template <typename T>
void __print(const T &x) {
  int f = 0;
  for (auto &i : x) cout << (f++ ? "," : ""), __print(i);
}
void _print() { cout << "]\n"; }
template <typename T, typename... V>
void _print(T t, V... v) {
  __print(t);
  if (sizeof...(v)) cout << ", ";
  _print(v...);
}
const int N = 1e5 + 7;
const int MOD = 1e9 + 7;
long long fastpow(long long b, long long p) {
  long long temp;
  if (p == 0) return 1;
  if (p == 1) return b;
  temp = fastpow(b, p / 2);
  temp = temp * temp % MOD;
  if (p & 1) {
    temp = temp * b % MOD;
  }
  return temp;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  long long u, v;
  cin >> u >> v;
  if (v < u || (v - u) & 1) {
    cout << "-1";
    return 0;
  } else if (v == u) {
    if (u == 0) {
      cout << 0;
      return 0;
    } else {
      cout << 1 << '\n';
      cout << v;
      return 0;
    }
  } else {
    long long temp = (v - u) / 2;
    if (u & temp)
      cout << 3 << '\n' << u << ' ' << temp << ' ' << temp;
    else
      cout << 2 << '\n' << (u ^ temp) << ' ' << temp;
  }
  return 0;
}
