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;
using DO = double;
using LL = long long;
using VI = vector<int>;
using pii = pair<int, int>;
const int mod = 1e9 + 7;
int power(int a, int b, int c = mod) {
  int ans = 1;
  for (; b; b >>= 1, a = (LL)a * a % c) {
    if (b & 1) ans = (LL)ans * a % c;
  }
  return ans;
}
int a[100];
int main() {
  LL u, v;
  cin >> u >> v;
  if (u > v) {
    puts("-1");
    return 0;
  }
  if (u % 2 != v % 2) {
    puts("-1");
    return 0;
  }
  for (int i = 0; i <= 62; i++) {
    LL s = 1LL << i;
    if (u & s) a[i] = 1;
  }
  v -= u;
  v /= 2;
  for (int i = 0; i <= 62; i++) {
    LL s = 1LL << i;
    if (v & s) a[i] += 2;
  }
  int m = 0;
  for (int i = 0; i <= 62; i++) m = max(m, a[i]);
  cout << m << endl;
  for (int i = 1; i <= m; i++) {
    LL d = 0;
    for (int j = 0; j <= 62; j++) {
      if (!a[j]) continue;
      a[j]--;
      d += 1LL << j;
    }
    cout << d;
    if (i < m) cout << ' ';
  }
  return 0;
}