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;
long long u, v;
long long uu[105] = {0}, vv[105] = {0};
long long ans[105] = {0};
int pu = 0, pv = 0;
int main() {
  long long rem;
  cin >> u >> v;
  if (u == v) {
    if (u == 0)
      cout << 0 << endl;
    else
      cout << 1 << endl << u << endl;
    return 0;
  }
  rem = u;
  while (rem) {
    uu[++pu] = rem % 2;
    rem >>= 1;
  }
  while (v) {
    vv[++pv] = v % 2;
    v >>= 1;
  }
  if (uu[1] != vv[1]) {
    cout << -1 << endl;
    return 0;
  }
  long long jw = 0;
  for (int i = 1; i <= max(pu, pv) + 5; ++i) {
    long long tmp = jw + uu[i];
    if (uu[i + 1] == vv[i + 1]) {
      if (((tmp + 2) / 2) % 2 == 1)
        ans[i] = 0;
      else
        ans[i] = 1;
    } else {
      if (((tmp + 2) / 2) % 2 == 1)
        ans[i] = 1;
      else
        ans[i] = 0;
    }
    jw = (tmp + ans[i] + ans[i]) / 2;
  }
  if (jw != 0) {
    cout << -1 << endl;
    return 0;
  }
  long long xx = 0, kk = 1;
  for (long long i = 1; i <= (long long)(max(pu, pv)); ++i) {
    xx += ans[i] * kk;
    kk *= 2;
  }
  if (xx + u == (xx ^ u)) {
    cout << 2 << endl << u + xx << ' ' << xx << endl;
    return 0;
  }
  cout << 3 << endl << u << ' ' << xx << ' ' << xx << endl;
  return 0;
}