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 N = 2e5 + 5;
const int M = 2e3 + 5;
const int mod = 998244353;
const double eps = 1e-8;
const double pi = acos(-1);
template <class T>
inline void read(T &x) {
  char c;
  x = 1;
  while ((c = getchar()) < '0' || c > '9')
    if (c == '-') x = -1;
  T res = c - '0';
  while ((c = getchar()) >= '0' && c <= '9') res = res * 10 + c - '0';
  x *= res;
}
int n, m;
char str[N];
int f[N];
void solve(int cas) {
  long long a, b;
  read(a);
  read(b);
  if (a == 0 && b == 0) {
    cout << 0 << endl;
    return;
  }
  if (a > b) {
    cout << -1 << endl;
    return;
  }
  if (a == b) {
    cout << 1 << endl << a << endl;
    return;
  }
  long long tmp = b - a;
  if (tmp & 1) {
    cout << -1 << endl;
    return;
  }
  long long ans1 = tmp >> 1;
  if ((ans1 + a) == (ans1 ^ a)) {
    cout << 2 << endl << (ans1 + a) << ' ' << ans1 << endl;
  } else {
    cout << 3 << endl << ans1 << ' ' << ans1 << ' ' << a << endl;
  }
}
int main() {
  int t;
  t = 1;
  for (int i = 1; i <= t; i++) {
    solve(i);
  }
  return 0;
}