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>
#pragma GCC target("avx2")
#pragma GCC optimization("O3")
#pragma GCC optimization("unroll-loops")
using namespace std;
vector<int64_t> poss(int64_t u, int64_t v) {
  int64_t i, j, k;
  j = (v - u) / 2;
  vector<int64_t> ans = {0, 0, 0};
  for (i = 62; i >= 0; i--) {
    int64_t x1, y1;
    x1 = y1 = 0;
    if ((1LL << i) & j) x1 = 1;
    if ((1LL << i) & u) y1 = 1;
    if (x1 == 1 && y1 == 1) return ans;
    if (x1 == 1 && y1 == 0) {
      ans[1] += (1LL << i);
      ans[2] += (1LL << i);
      continue;
    }
    if (x1 == 0 && y1 == 1) {
      ans[1] += (1LL << i);
      continue;
    }
  }
  ans[0] = 1;
  return ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int64_t i, j, k, n, u, v;
  cin >> u >> v;
  if (((u & 1) != (v & 1)) || (u > v)) {
    cout << "-1\n";
    return 0;
  }
  if (u == 0 && v == 0) {
    cout << "0\n";
    return 0;
  }
  if (u == v) {
    cout << "1\n" << u << "\n";
    return 0;
  }
  vector<int64_t> arr = poss(u, v);
  if (arr[0] == 1) {
    cout << "2\n";
    cout << arr[1] << " " << arr[2] << "\n";
  } else {
    cout << "3\n";
    cout << u << " " << (v - u) / 2 << " " << (v - u) / 2 << "\n";
  }
  return 0;
}
