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 quickly(long long a, long long b) {
  long long t1 = 1, t2 = a;
  while (b) {
    if (b % 2 == 1) {
      t1 = t1 * t2 % 998244353;
    }
    t2 = t2 * t2 % 998244353;
    b = b / 2;
  }
  return t1;
}
long long t, x, a, b, c, i, j, k, n, m, arr[200005];
int main() {
  cin >> a >> b;
  c = a;
  i = 0;
  while (c) {
    if (c % 2 == 1) arr[i]++;
    c = c / 2;
    i++;
  }
  b = b - a;
  n = i;
  long long tem = b;
  if (b < 0 || b % 2 == 1)
    cout << -1;
  else {
    b = b / 2;
    i = 0;
    while (b) {
      if (b % 2 == 1) {
        arr[i] += 2;
      }
      i++;
      b = b / 2;
    }
    n = i > n ? i : n;
    m = -1;
    if (a != 0 || tem != 0) {
      for (i = 0; i < n; i++) {
        m = m > arr[i] ? m : arr[i];
      }
      cout << m << endl;
      m = -1;
      while (m != 0) {
        a = 0;
        j = 1;
        m = 0;
        for (i = 0; i < n; i++) {
          m = m > arr[i] ? m : arr[i];
          if (arr[i] > 0) {
            a = a + j;
            arr[i]--;
          }
          j = j * 2;
        }
        if (m != 0) cout << a << ' ';
      }
    } else
      cout << 0;
  }
}
