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;
const int N = 105;
long long n, m;
int f[N][N * 2][N];
pair<int, pair<int, int> > trace[N][N * 2][N];
int main() {
  if (fopen("A.inp", "r"))
    freopen("A.inp", "r", stdin), freopen("A.out", "w", stdout);
  cin >> n >> m;
  if (!n && !m) {
    cout << 0;
    return 0;
  }
  int nbit = 1;
  for (; 1ll << nbit <= m; nbit++)
    ;
  if (1ll << nbit <= n) {
    cout << -1;
    return 0;
  }
  f[0][0][0] = 1;
  for (int i = 0; i < nbit; i++)
    for (int j = 0; j <= 100 * 2; j++)
      for (int k = 0; k <= 100; k++)
        if (f[i][j][k]) {
          if ((j / 2 + (n >> i & 1ll)) % 2 != (m >> i & 1ll)) continue;
          for (int used = (n >> i & 1ll); used <= 100; used += 2) {
            f[i + 1][j / 2 + used][max(k, used)] = 1;
            trace[i + 1][j / 2 + used][max(k, used)] =
                make_pair(used, pair<int, int>(j, k));
          }
        }
  for (int i = 1; i <= 100; i++)
    if (f[nbit][1][i]) {
      vector<long long> res(i, 0);
      int j = 1, k = i;
      for (int ii = nbit; ii >= 1; ii--) {
        int cnt = 0;
        while (trace[ii][j][k].first--) res[cnt++] += (1ll << (ii - 1));
        pair<int, int> tmp = trace[ii][j][k].second;
        j = tmp.first;
        k = tmp.second;
      }
      cout << (int)(res.size()) << endl;
      for (long long x : res) cout << x << " ";
      return 0;
    }
  cout << -1;
}
