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;
long long int max(long long int a, long long int b) {
  if (a > b) {
    return a;
  }
  return b;
}
long long int gcd(long long int a, long long int b) {
  if (a == 0) {
    return b;
  }
  return gcd(b % a, a);
}
long long int min(long long int a, long long int b) {
  if (a < b) {
    return a;
  }
  return b;
}
long long int power(long long int x, long long int n) {
  long long int res = 1;
  while (n > 0) {
    if (n % 2 == 1) {
      res = (res * x);
    }
    x = (x * x);
    n = n / 2;
  }
  return res;
}
long long int factorialNumInverse[1000001];
long long int naturalNumInverse[1000001];
long long int fact[1000001];
void InverseofNumber(long long int n, long long int p) {
  naturalNumInverse[0] = naturalNumInverse[1] = 1;
  for (long long int i = 2; i <= n; i++)
    naturalNumInverse[i] = naturalNumInverse[p % i] * (p - p / i) % p;
}
void InverseofFactorial(long long int n, long long int p) {
  factorialNumInverse[0] = factorialNumInverse[1] = 1;
  for (long long int i = 2; i <= n; i++)
    factorialNumInverse[i] =
        (naturalNumInverse[i] * factorialNumInverse[i - 1]) % p;
}
void factorial(long long int n, long long int p) {
  fact[0] = 1;
  for (long long int i = 1; i <= n; i++) {
    fact[i] = (fact[i - 1] * i) % p;
  }
}
long long int nCrModp(long long int n, long long int r, long long int p) {
  long long int ans =
      ((fact[n] * factorialNumInverse[r]) % p * factorialNumInverse[n - r]) % p;
  return ans;
}
long long int n;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  factorial(200005, 1000000007);
  InverseofNumber(200005, 1000000007);
  InverseofFactorial(200005, 1000000007);
  long long int u, v;
  cin >> u >> v;
  if (u == 0 && v == 0) {
    cout << 0 << "\n";
    return 0;
  }
  if (u > v || u % 2 != v % 2) {
    cout << -1 << "\n";
    return 0;
  }
  if (u == v && v != 0) {
    cout << 1 << "\n";
    cout << u << "\n";
    return 0;
  }
  long long int x = (v - u) / 2;
  if ((x & u) == 0) {
    cout << 2 << "\n";
    cout << u + x << " " << x << "\n";
  } else {
    cout << 3 << "\n";
    cout << u << " " << x << " " << x << "\n";
  }
  return 0;
}