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;
void comp(long long x, vector<long long>& arr) {
  for (long long i = 0; i < 64; i++) {
    if (x & (1ll << i))
      arr[i] = 1;
    else
      arr[i] = 0;
  }
}
int main() {
  long long u, v, i, j;
  cin >> u >> v;
  vector<long long> cntu(64), cntv(64);
  vector<long long> ans(64);
  comp(u, cntu);
  comp(v, cntv);
  long long cnt = 0;
  bool poss = true;
  for (i = 63; i >= 0; i--) {
    ans[i] += cntv[i];
    if (cntu[i] == 0 && ans[i] % 2 != 0 || (cntu[i] == 1 && ans[i] % 2 == 0)) {
      if (ans[i] == 0) {
        for (j = i + 1; j < 64; j++)
          if (ans[j] >= 2) break;
        if (j < 64) {
          while (j > i) {
            ans[j] -= 2;
            ans[j - 1] += 4;
            j--;
          }
        }
      }
      if (ans[i] == 0 || i == 0)
        poss = false;
      else {
        ans[i - 1] += 2;
        ans[i]--;
      }
    }
  }
  if (!poss) {
    cout << "-1";
    return 0;
  }
  long long n = 0;
  for (i = 0; i < 64; i++) n = max(n, ans[i]);
  vector<long long> arr(n);
  for (i = 0; i < n; i++) {
    for (j = 0; j < 64; j++) {
      if (i < ans[j]) arr[i] |= (1ll << j);
    }
  }
  cout << n << endl;
  for (i = 0; i < n; i++) cout << arr[i] << " ";
}