#include <bits/stdc++.h>
using namespace std;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  long long _and, _xor, sum, ans = 1, zero = 0;
  cin >> sum >> _xor;
  _and = (sum - _xor) / 2;
  if ((sum - _xor) % 2 || sum < _xor || _xor & _and) return cout << 0, 0;
  if (sum == _xor) zero = 2;
  while (_xor) {
    if (_xor % 2) ans *= 2;
    _xor /= 2;
  }
  cout << ans - zero;
}

### Prompt

In cpp, your task is to solve the following problem:

Two positive integers a and b have a sum of s and a bitwise XOR of x. How many possible values are there for the ordered pair (a, b)?

Input

The first line of the input contains two integers s and x (2 ≤ s ≤ 1012, 0 ≤ x ≤ 1012), the sum and bitwise xor of the pair of positive integers, respectively.

Output

Print a single integer, the number of solutions to the given conditions. If no solutions exist, print 0.

Examples

Input

9 5


Output

4


Input

3 3


Output

2


Input

5 2


Output

0

Note

In the first sample, we have the following solutions: (2, 7), (3, 6), (6, 3), (7, 2).

In the second sample, the only solutions are (1, 2) and (2, 1).

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  long long _and, _xor, sum, ans = 1, zero = 0;
  cin >> sum >> _xor;
  _and = (sum - _xor) / 2;
  if ((sum - _xor) % 2 || sum < _xor || _xor & _and) return cout << 0, 0;
  if (sum == _xor) zero = 2;
  while (_xor) {
    if (_xor % 2) ans *= 2;
    _xor /= 2;
  }
  cout << ans - zero;
}

```