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 maxn = 1e5 + 20, inf = 3e5 + 5;
long long t2[60];
void pre() {
  t2[0] = 1;
  for (int i = 1; i < 60; i++) {
    t2[i] = t2[i - 1] * 2;
  }
}
int main() {
  pre();
  long long x, y, z, tt;
  cin >> x >> y;
  tt = x;
  if (x > y || (x % 2 != y % 2)) {
    cout << "-1";
    return 0;
  } else if (x == y) {
    if (x == 0) {
      cout << 0;
      return 0;
    }
    cout << 1 << endl;
    cout << x << endl;
    return 0;
  }
  z = y - x;
  bool mark = 1;
  vector<int> v, vv;
  for (int i = 59; i > -1; i--) {
    if (z >= t2[i]) {
      v.push_back(i - 1);
      z -= t2[i];
    }
  }
  for (int i = 59; i > -1; i--) {
    if (tt >= t2[i]) {
      tt -= t2[i];
    } else
      vv.push_back(i);
  }
  for (int i = 0; i < v.size(); i++) {
    bool po = 0;
    for (int j = 0; j < vv.size(); j++) {
      if (v[i] == vv[j]) po = 1;
    }
    if (!po) {
      mark = 0;
      break;
    }
  }
  z = 0;
  y = 0;
  if (mark) {
    for (int i : v) {
      x += t2[i];
      z += t2[i];
    }
    cout << 2 << endl;
    cout << x << " " << z;
  } else {
    for (int i : v) {
      y += t2[i];
      z += t2[i];
    }
    cout << 3 << endl;
    cout << x << " " << y << " " << z << endl;
  }
  return 0;
}
