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;
bool isFile = 0;
const long long INF = 1e18 + 1234;
const int N = 1e5 + 123;
const long long MOD = 998244353;
void faster() {
  if (isFile) {
    freopen(
        ""
        ".in",
        "r", stdin);
    freopen(
        ""
        ".out",
        "w", stdout);
  }
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
}
int a[N + 1];
int g[N][2];
void solve() {
  long long u, v;
  cin >> u >> v;
  if (u % 2 != v % 2 || u > v)
    cout << -1;
  else if (u == v) {
    if (!u)
      cout << 0;
    else {
      cout << 1;
      cout << "\n";
      cout << u;
    }
  } else {
    long long p = (v - u) / 2;
    long long rp = v - 2 * p;
    if ((rp ^ p) == p + rp) {
      cout << 2;
      cout << "\n";
      cout << rp + p << ' ' << p;
    } else {
      cout << 3;
      cout << "\n";
      cout << rp << ' ' << p << ' ' << p;
    }
  }
}
int main() {
  faster();
  int t = 1;
  while (t--) {
    solve();
  }
  return 0;
}
