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 power(long long a, long long b) {
  long long ans = 1;
  while (b > 0) {
    if (b & 1) {
      ans = (ans * a);
    }
    a = (a * a);
    b >>= 1;
  }
  return ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long a, b, i, l = 0, u;
  cin >> a >> b;
  if (a > b) {
    cout << "-1";
    return 0;
  }
  if (a == 0 && b == 0) {
    cout << "0";
    return 0;
  }
  if (a == 0 && b % 2 == 0) {
    cout << "2\n" << b / 2 << " " << b / 2;
    return 0;
  }
  if (a == b) {
    cout << 1 << "\n" << a;
    return 0;
  }
  if (a % 2 != 0 && b % 2 != 0) {
    long long d = b - a;
    if (d == 0) {
      cout << "1\n" << 1;
      return 0;
    }
    if ((a & (d / 2)) == 0) {
      cout << "2"
           << "\n";
      cout << a + d / 2 << " ";
      cout << d / 2;
      return 0;
    }
    cout << "3"
         << "\n";
    cout << a << " ";
    cout << d / 2 << " " << d / 2;
    return 0;
  }
  if (a % 2 == 0 && b % 2 == 0) {
    long long d = a;
    vector<int> v;
    while (d > 0) {
      v.push_back(d & 1);
      d = d >> 1;
    }
    while (v.size() <= 60) {
      v.push_back(0);
    }
    for (i = 60; i >= 0; i--) {
      if (v[i] == 0) {
        u = power(2, i);
        if (a + l + 2 * u == b) break;
        if (a + l + 2 * u < b) {
          l = l + 2 * u;
        }
      }
    }
    if (((a + l / 2 + u) ^ (l / 2 + u)) != a) {
      long long d = b - a;
      cout << "3"
           << "\n";
      cout << a << " ";
      cout << d / 2 << " " << d / 2;
      return 0;
    }
    cout << "2\n";
    cout << a + l / 2 + u << " " << l / 2 + u << "\n";
    return 0;
  }
  cout << "-1";
  return 0;
}