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;
const int N = 1000010;
vector<long long> is_prime(N + 1, true);
long long to_int(string s) {
  stringstream geek(s);
  long long x = 0;
  geek >> x;
  return x;
}
string to_string(long long n) {
  ostringstream str1;
  str1 << n;
  string geek = str1.str();
  return geek;
}
bool sortinrev(const pair<int, int> &a, const pair<int, int> &b) {
  return (a.first > b.first);
}
bool sortbysecdesc(const pair<int, int> &a, const pair<int, int> &b) {
  return a.second > b.second;
}
bool sortbysec(const pair<int, int> &a, const pair<int, int> &b) {
  return (a.second < b.second);
}
long long powerm(long long a, long long k) {
  if (!k) {
    return 1;
  }
  long long b = powerm(a, k / 2);
  b = b * b % (1000000000 + 7);
  if (k % 2) {
    return a * b % (1000000000 + 7);
  } else {
    return b;
  }
}
long long power(long long a, long long b) {
  if (b == 1) return a;
  if (b == 0) return 1;
  long long m1 = power(a, b / 2);
  if (b % 2) return m1 * m1 * a;
  return m1 * m1;
}
void seive_of_eras(long long n) {
  is_prime[0] = 0;
  is_prime[1] = 0;
  for (long long i = 2; i * i <= n; i++) {
    if (is_prime[i]) {
      for (long long j = i * i; j <= n; j += i) {
        is_prime[j] = false;
      }
    }
  }
}
bool isprime(long long a) {
  if (a <= 1) return false;
  if (a == 2 | a == 3) return true;
  if (a % 2 == 0 || a % 3 == 0) return false;
  for (long long i = 5; i * i <= a; i = i + 6) {
    if (a % i == 0 || a % (i + 2) == 0) return false;
  }
  return true;
}
void codeit() {
  long long u, v;
  cin >> u >> v;
  if (u == v) {
    if (!u)
      printf("0");
    else {
      cout << 1 << endl;
      cout << u << endl;
    }
    return;
  }
  if (u > v || u % 2 != v % 2) {
    cout << -1 << endl;
    return;
  }
  long long x = (v - u) / 2;
  if ((u & x)) {
    cout << 3 << endl;
    cout << u << " " << x << " " << x << endl;
    return;
  } else {
    cout << 2 << endl;
    cout << (u ^ x) << " " << x << endl;
  }
  return;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(0);
  long long tt;
  tt = 1;
  for (long long i = 0; i < tt; i++) {
    codeit();
  }
}