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 int maxn = 4e6 + 7;
long long u, v;
long long ans[5];
void solve() {
  scanf("%lld%lld", &u, &v);
  if (u > v) {
    puts("-1");
    return;
  }
  if (u == v) {
    if (u == 0)
      printf("0");
    else
      printf("1\n%lld", u);
    return;
  }
  long long d = v - u;
  bool flag = false;
  for (long long i = 60; i >= 0; i--) {
    if ((1LL << i) & u) {
      ans[0] += (1LL << i);
    }
    if ((1LL << i) & d) {
      if (i == 0) {
        puts("-1");
        return;
      }
      if ((1LL << (i - 1)) & u) {
        for (int j = (1); j <= (2); j++)
          ans[j] += (1LL << (i - 1)), flag = true;
      } else {
        for (int j = (0); j <= (1); j++) {
          ans[j] += (1LL << (i - 1));
        }
      }
    }
  }
  int num = flag ? 2 : 1;
  printf("%d\n", num + 1);
  for (int i = (0); i <= (num); i++) printf("%lld ", ans[i]);
  puts("");
}
int main() { solve(); }