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 long double PI = (long double)(3.1415926535897932384626433832795);
int parity_check(int n) {
  int par = n;
  par = par ^ (par >> 1);
  par = par ^ (par >> 2);
  par = par ^ (par >> 4);
  par = par ^ (par >> 8);
  par = par ^ (par >> 16);
  if (par % 2 == 1) {
    return 1;
  } else {
    return 0;
  }
}
int lcs(string X, string Y, int m, int n) {
  int L[m + 1][n + 1];
  int i, j;
  for (i = 0; i <= m; i++) {
    for (j = 0; j <= n; j++) {
      if (i == 0 || j == 0)
        L[i][j] = 0;
      else if (X[i - 1] == Y[j - 1] && i != j)
        L[i][j] = L[i - 1][j - 1] + 1;
      else
        L[i][j] = max(L[i - 1][j], L[i][j - 1]);
    }
  }
  return L[m][n];
}
int long long gcd(int long long a, int long long b) {
  if (a == 0) {
    return b;
  }
  return gcd(b % a, a);
}
int long long lcm(int long long a, int long long b) {
  int long long pro = a * b;
  int long long ans = gcd(a, b);
  pro = pro / ans;
  return pro;
}
int long long fast_power(int long long a, int long long b, int long long p) {
  if (b == 0) {
    return (1 % p);
  }
  int long long t = fast_power(a, b / 2, p);
  int long long ans = (t * t) % p;
  if (b % 2) {
    ans = (ans * (a % p)) % p;
  }
  return ans;
}
bool prime_check(int long long n) {
  int long long i, j;
  if (n == 1) {
    return false;
  }
  for (i = 2; i <= sqrt(n); i++) {
    if (n % i == 0) {
      return false;
    }
  }
  return true;
}
int long long fact(int long long n, int long long mod) {
  int long long prod = 1;
  for (int long long i = 1; i <= n; i++) {
    prod = (prod * i) % mod;
  }
  return prod;
}
void mera_sieve() {
  bool prime[6000013];
  memset(prime, true, sizeof(prime));
  for (int long long p = 2; p * p <= 6000012; p++) {
    if (prime[p] == true) {
      for (int long long i = p * p; i <= 6000012; i += p) prime[i] = false;
    }
  }
  for (int long long i = 2; i <= 6000012; i++) {
    if (prime[i]) {
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  int long long xo, sum, i, j;
  cin >> xo >> sum;
  if (xo > sum || (sum - xo) % 2 == 1) {
    cout << "-1";
  } else {
    if (xo == sum) {
      if (sum != 0)
        cout << 1 << "\n" << xo;
      else
        cout << 0 << "\n";
    } else {
      int long long diff = sum - xo;
      diff /= 2;
      if ((xo & diff) == 0) {
        cout << 2 << "\n" << diff + xo << " " << diff;
      } else {
        cout << 3 << "\n" << xo << " " << diff << " " << diff;
      }
    }
  }
  return 0;
}