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;
long long int MOD = 998244353;
long long int MX = 1000000000;
long long int u, v;
vector<long long int> bin(long long int x) {
  vector<long long int> res;
  for (int i = 1; i <= 70; i++) {
    res.push_back(x % 2);
    x /= 2;
  }
  return res;
}
void prnt(vector<long long int> a) {
  for (int i = 0; i < a.size(); i++) cout << a[i];
  cout << endl;
}
void solve(long long int a, long long int b, long long int c) {
  vector<long long int> aa = bin(a);
  vector<long long int> bb = bin(b);
  vector<long long int> cc = bin(c);
  vector<long long int> dd(70, 0);
  bool three = 0;
  for (int i = 0; i < 70; i++) {
    dd[i] = aa[i] + bb[i] + cc[i];
    if (dd[i] == 3) three = 1;
  }
  if (three) {
    cout << 3 << endl << a << " " << b << " " << c << endl;
  } else {
    long long int x = 0, y = 0;
    for (long long int i = 0; i < 70; i++) {
      long long int adda = (((long long int)1) << i);
      if (dd[i] == 2) {
        x = x + adda;
        y = y + adda;
      } else if (dd[i] == 1) {
        x = x + adda;
      }
    }
    cout << 2 << endl << x << " " << y << endl;
  }
}
int main() {
  ios::sync_with_stdio(false), cin.tie(NULL), cout.tie(NULL);
  cin >> u >> v;
  if (u > v)
    cout << "-1" << endl;
  else if (u == v) {
    if (u == 0)
      cout << 0 << endl;
    else
      cout << "1\n" << u << endl;
  } else {
    long long int x = v - u;
    if (x % 2 != 0)
      cout << -1 << endl;
    else {
      long long int a = u;
      long long int b = x / 2;
      long long int c = x / 2;
      solve(a, b, c);
    }
  }
  return 0;
}
