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;
int power(int a, int b, int m, int ans = 1) {
  for (; b; b >>= 1, a = 1LL * a * a % m)
    if (b & 1) ans = 1LL * ans * a % m;
  return ans;
}
int a[100];
int main() {
  long long u, v;
  cin >> v >> u;
  if (u == 0 and v == 0) {
    puts("0");
    return 0;
  }
  if (v > u or (v - u) % 2 != 0) {
    puts("-1");
    return 0;
  }
  if (u == v) {
    puts("1");
    printf("%I64d\n", u);
    return 0;
  }
  int len = 0;
  long long x = u, y = v;
  while (v) {
    if (u % 2 != v % 2) {
      printf("3\n");
      printf("%I64d %I64d %I64d\n", (x - y) / 2, (x - y) / 2, y);
      return 0;
    }
    if (v % 2 == 0) {
      if ((v / 2) % 2 == (u / 2) % 2) {
        a[len++] = 0;
        v /= 2;
        u /= 2;
      } else {
        a[len++] = 1;
        v /= 2;
        u /= 2;
        u--;
      }
    } else {
      a[len++] = 1;
      u /= 2;
      v /= 2;
    }
  }
  if (u % 2) {
    printf("3\n");
    printf("%I64d %I64d %I64d\n", (x - y) / 2, (x - y) / 2, y);
    return 0;
  }
  u /= 2;
  long long ans = 0, tmp = 1;
  for (int i = 0; i < len; i++) {
    ans += a[i] * tmp;
    tmp *= 2;
  }
  ans += tmp * u;
  printf("2\n");
  printf("%I64d %I64d\n", ans, x - ans);
  return 0;
}
