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;
int BIT[65];
int64_t u, v;
int64_t POW(int x) {
  if (x == 0) return 1;
  int64_t tg = POW(x / 2);
  tg = tg * tg;
  if (x % 2 == 1) tg = tg * 2;
  return tg;
}
int Get(int64_t tt, int x) { return (tt >> x) & 1; }
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cin >> u >> v;
  if (u > v) {
    cout << -1;
    return 0;
  }
  for (int i = 0; i <= 60; ++i) {
    if (Get(u, i) == 1) {
      ++BIT[i];
      v = v - POW(i);
    }
  }
  if (v % 2 == 1) {
    cout << -1;
    return 0;
  }
  for (int i = 1; i <= 60; ++i) {
    if (v == 0) break;
    int64_t s = v % POW(i + 1);
    if (s % POW(i) != 0) {
      cout << -1;
      return 0;
    }
    v = v - s;
    s = s / POW(i);
    if (s % 2 == 1) {
      --s;
      BIT[i - 1] += 2;
    }
    BIT[i] += s;
  }
  if (v > 0) {
    cout << -1;
    return 0;
  }
  int rs = 0;
  for (int i = 0; i <= 60; ++i) rs = max(rs, BIT[i]);
  cout << rs << '\n';
  for (int i = 1; i <= rs; ++i) {
    int64_t t = 0;
    for (int j = 0; j <= 60; ++j)
      if (BIT[j] > 0) {
        --BIT[j];
        t = t + POW(j);
      }
    cout << t << " ";
  }
  return 0;
}
