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;
inline void normal(long long int &a) { a = (a + 998244353) % 998244353; }
inline int modMul(long long int a, long long int b) {
  a %= 998244353, b %= 998244353;
  normal(a), normal(b);
  return (a * b) % 998244353;
}
inline int modAdd(long long int a, long long int b) {
  a %= 998244353, b %= 998244353;
  normal(a), normal(b);
  return (a + b) % 998244353;
}
inline int modSub(long long int a, long long int b) {
  a %= 998244353, b %= 998244353;
  normal(a), normal(b);
  a -= b;
  normal(a);
  return a;
}
inline int modPow(long long int b, long long int p) {
  int r = 1;
  while (p) {
    if (p & 1) r = modMul(r, b);
    b = modMul(b, b);
    p >>= 1;
  }
  return r;
}
inline int modInv(int a) { return modPow(a, 998244353 - 2); }
inline int modDiv(int a, int b) { return modMul(a, modInv(b)); }
inline int fact(int r) {
  int ans = 1;
  while (r) ans = modMul(ans, r), r--;
  return ans;
}
inline int invfact(int r) {
  int ans = 1;
  while (r) ans = modMul(ans, modInv(r)), r--;
  return ans;
}
inline double dgcd(double a, double b) {
  if (a < b) return dgcd(b, a);
  if (b < .0001) return a;
  return dgcd(b, fmod(a, b));
}
const long long int N = 305;
const long long int INF = (long long int)1e12;
long long int power(long long int base, long long int exp) {
  long long int res = 1;
  while (exp > 0) {
    if (exp % 2 == 1) res = (res * base);
    base = (base * base);
    exp /= 2;
  }
  return res;
}
bool isPrime(int n) {
  if (n <= 1) return false;
  if (n <= 3) return true;
  if (n % 2 == 0 || n % 3 == 0) return false;
  for (int i = 5; i * i <= n; i = i + 6)
    if (n % i == 0 || n % (i + 2) == 0) return false;
  return true;
}
void Abesaale() {
  long long int u, v;
  cin >> u >> v;
  if (u > v)
    cout << -1;
  else {
    if ((v - u) % 2 != 0)
      cout << -1;
    else {
      long long int k1 = (v - u) / 2;
      if ((k1 + u) == (k1 ^ u)) {
        if (k1 == 0 && u == 0) {
          cout << 0 << endl;
          return;
        } else if (k1 == 0) {
          cout << 1 << endl;
          cout << v << endl;
          return;
        }
        cout << 2 << endl;
        cout << k1 + u << " " << k1;
      } else {
        cout << 3 << endl;
        cout << k1 << " " << k1 << " " << u << endl;
      }
    }
  }
  return;
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int t = 1;
  clock_t start, end;
  start = clock();
  cout << fixed << setprecision(11);
  while (t--) {
    Abesaale();
  }
  end = clock();
  double time_taken = double(end - start) / double(CLOCKS_PER_SEC);
}