#include <bits/stdc++.h>
using namespace std;
template <class T>
inline istream& operator>>(istream& str, vector<T>& a) {
  for (auto& i : a) str >> i;
  return str;
}
long long M = 1e9 + 7;
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  long long int second, x;
  cin >> second >> x;
  long long int a = second - x;
  if (a % 2) {
    cout << 0 << "\n";
    return 0;
  }
  a /= 2;
  long long int first = 0;
  long long int ans = 1;
  for (long long int i = 0; i < 64; i++) {
    long long int xx = x & (1ll << i);
    long long int aa = a & (1ll << i);
    if (xx && aa) {
      cout << 0 << "\n";
      return 0;
    }
    if (xx && !aa) ans = ans * 2ll;
  }
  if (a == 0) ans -= 2;
  cout << ans << "\n";
}

### Prompt

Please provide a CPP coded solution to the problem described below:

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;
template <class T>
inline istream& operator>>(istream& str, vector<T>& a) {
  for (auto& i : a) str >> i;
  return str;
}
long long M = 1e9 + 7;
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  long long int second, x;
  cin >> second >> x;
  long long int a = second - x;
  if (a % 2) {
    cout << 0 << "\n";
    return 0;
  }
  a /= 2;
  long long int first = 0;
  long long int ans = 1;
  for (long long int i = 0; i < 64; i++) {
    long long int xx = x & (1ll << i);
    long long int aa = a & (1ll << i);
    if (xx && aa) {
      cout << 0 << "\n";
      return 0;
    }
    if (xx && !aa) ans = ans * 2ll;
  }
  if (a == 0) ans -= 2;
  cout << ans << "\n";
}

```