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 gcd(long long a, long long b) {
  if (a == 0) return b;
  return gcd(b % a, a);
}
bool isperfect(long long n) {
  long long y = sqrt(n);
  if (n % y == 0 && y * y == n) return true;
  return false;
}
bool comp(long long a, long long b) { return a > b; }
long long powi(long long a, long long b) {
  long long ans = 1;
  while (b > 0) {
    if (b & 1) ans = ans * a;
    b = b >> 1;
    a = a * a;
  }
  return ans;
}
bool isprime(long long n) {
  for (int i = 2; i * i <= n; i++) {
    if (n % i == 0) return false;
  }
  return true;
}
bool prime[10000000];
void seive() {
  prime[1] = true;
  for (int i = 2; i * i < 10000000; i++) {
    if (prime[i]) continue;
    for (int j = i * i; j < 10000000; j += i) {
      prime[j] = true;
    }
  }
}
int kmp(string t) {
  int n = t.length();
  int kmp[n];
  kmp[0] = 0;
  int i = 0, r = 1;
  while (r < n) {
    if (t[r] == t[i]) {
      kmp[r] = i + 1;
      i++;
      r++;
    } else {
      if (i == 0) {
        kmp[r] = 0;
        r++;
      } else
        i = kmp[i - 1];
    }
  }
  return kmp[n - 1];
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long xorr, sum;
  cin >> xorr >> sum;
  if (xorr > sum || (sum - xorr) % 2) {
    cout << "-1";
    return 0;
  }
  if ((sum == xorr && sum == 0)) {
    cout << "0";
    return 0;
  }
  if (sum == xorr) {
    cout << "1" << endl << sum;
    return 0;
  }
  long long andi = (sum - xorr) / 2;
  long long nandi = (sum + xorr) / 2;
  if ((andi ^ nandi) == xorr) {
    cout << 2 << endl;
    cout << andi << " " << nandi;
  } else {
    cout << "3" << endl;
    cout << andi << " " << andi << " " << xorr;
  }
}
