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>
#pragma GCC optimize(2)
using namespace std;
inline long long read() {
  long long s = 0, w = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') w = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') s = s * 10 + ch - '0', ch = getchar();
  return s * w;
}
void put1() { puts("YES"); }
void put2() { puts("NO"); }
void put3() { puts("-1"); }
long long qp(long long a, long long b, long long p) {
  long long ans = 1;
  while (b) {
    if (b & 1) {
      ans = (ans * a) % p;
      --b;
    }
    a = (a * a) % p;
    b >>= 1;
  }
  return ans % p;
}
long long Inv(long long x, long long p) { return qp(x, p - 2, p); }
long long Cal(long long n, long long m, long long p) {
  if (m > n) return 0;
  long long ans = 1;
  for (int i = 1; i <= m; ++i) ans = ans * Inv(i, p) % p * (n - i + 1) % p;
  return ans % p;
}
const int manx = 2e5 + 5;
int main() {
  long long u = read(), v = read();
  if (u > v || ((v - u) & 1)) {
    put3();
    return 0;
  } else if (u == v) {
    if (!u)
      cout << 0;
    else
      cout << 1 << " " << u;
    return 0;
  }
  long long x = (v - u) / 2, y = (v + u) / 2;
  if (((x ^ y) == u) && x + y == v)
    cout << 2 << endl << x << " " << y << endl;
  else {
    cout << 3 << endl;
    cout << u << " " << x << " " << x << endl;
  }
  return 0;
}