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 a[101000] = {0}, b[100] = {0};
int main() {
  long long i, u, v, ccc = v - u;
  scanf("%lld%lld", &u, &v);
  ccc = v - u;
  if (ccc < 0 || ccc % 2) {
    printf("-1");
    return 0;
  }
  if (u == v && u == 0) {
    printf("0");
    return 0;
  }
  if (u == v) {
    printf("1\n%lld", u);
    return 0;
  }
  long long te = 1, m, n, c, d, cc, ans1 = 0, ans2 = 0;
  for (i = 0; i <= 62; i++) {
    c = 0, d = 0;
    if ((te | u) == u) c = 1;
    if ((te | v) == v) d = 1;
    if (i > 0) {
      if (m == 0) {
        if (c != d) {
          ans1 += te / 2;
          ans2 += te / 2;
        }
      } else {
        ans2 += te / 2;
      }
    }
    te *= 2;
    m = c;
    n = d;
  }
  if (ans1 + ans2 == v && (ans1 ^ ans2) == u) {
    printf("2\n");
    printf("%lld %lld\n", ans1, ans2);
    return 0;
  }
  printf("3\n");
  printf("%lld %lld %lld\n", u, ccc / 2, ccc / 2);
}
