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 maxx = 1e5 + 10;
const long long mod = 1e9 + 7;
int ans[maxx], num[maxx];
int main() {
  long long u, v;
  cin >> u >> v;
  if (u > v || ((v - u) & 1)) return 0 * puts("-1");
  for (int i = 62; i >= 0; i--) {
    num[i] += ((v >> i) & 1);
    if ((u >> i) & 1) {
      if (num[i] & 1) {
        ;
      } else {
        if (num[i]) {
          if (i > 0) num[i - 1] += 2;
          num[i]--;
        } else {
          for (int j = i + 1; j <= 62; j++) {
            if (num[j] >= 2) {
              for (int k = j; k > i; k--) {
                num[k] -= 2;
                num[k - 1] += 4;
              }
              break;
            }
          }
          if (i > 0) num[i - 1] += 2;
          num[i]--;
        }
      }
    } else {
      if (num[i] & 1) {
        if (num[i]) {
          if (i > 0) num[i - 1] += 2;
          num[i]--;
        } else {
          for (int j = i + 1; j <= 62; j++) {
            if (num[j] >= 2) {
              for (int k = j; k > i; k--) {
                num[k] -= 2;
                num[k - 1] += 4;
              }
              break;
            }
          }
          if (i > 0) num[i - 1] += 2;
          num[i]--;
        }
      } else {
        ;
      }
    }
  }
  int n = 0;
  for (int i = 62; i >= 0; i--) {
    n = max(n, num[i]);
  }
  cout << n << endl;
  for (int i = 1; i <= n; i++) {
    long long x = 0;
    for (int j = 0; j <= 62; j++) {
      if (num[j]) {
        x += (1ll << j);
        num[j]--;
      }
    }
    printf("%lld ", x);
  }
}