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;
long long max(long long a, long long b) { return a > b ? a : b; }
vector<long long> conv(long long num) {
  vector<long long> ans(64, 0);
  for (int i = 0; i < 64; i++) {
    ans[i] = num % 2;
    num = num / 2;
  }
  return ans;
}
long long conv_back(vector<long long> b) {
  long long ans = b[63];
  for (int i = 62; i >= 0; i--) ans = ans * 2 + b[i];
  return ans;
}
int helper() {
  long long x, s;
  cin >> x >> s;
  if (x == 0) {
    if (s == 0) {
      cout << 0 << endl;
      return 0;
    } else if (s % 2 == 1) {
      cout << -1 << endl;
      return 0;
    } else {
      long long ans = s / 2;
      cout << 2 << endl << ans << " " << ans << endl;
      return 0;
    }
  }
  if (x == s) {
    cout << 1 << endl << x << endl;
    return 0;
  }
  if (x > s) {
    cout << -1 << endl;
    return 0;
  }
  if ((s - x) % 2 == 1) {
    cout << -1 << endl;
    return 0;
  } else {
    vector<long long> xb = conv(x);
    long long diff = (s - x) / 2;
    vector<long long> diffb = conv(diff);
    vector<long long> finalb(64, 0);
    int maxv = 0;
    for (int i = 0; i < 64; i++) {
      finalb[i] = xb[i] + 2 * diffb[i];
      maxv = max(maxv, finalb[i]);
    }
    if (maxv == 3) {
      cout << 3 << endl;
      cout << x << " " << diff << " " << diff << endl;
    } else {
      vector<long long> temp1(64, 0);
      for (int i = 0; i < 64; i++) {
        if (finalb[i] > 0) {
          temp1[i] = 1;
          finalb[i]--;
        }
      }
      long long num1 = conv_back(temp1);
      long long num2 = conv_back(finalb);
      cout << 2 << endl;
      cout << num1 << " " << num2 << endl;
    }
    return 0;
  }
  return 0;
}
int main() {
  helper();
  return 0;
}