#include <bits/stdc++.h>
using namespace std;
long long int countSetBits(long long int n) {
  if (n == 0)
    return 0;
  else
    return (n & 1) + countSetBits(n >> 1);
}
int main() {
  long long int s, x;
  cin >> s >> x;
  if (x > s || (s - x) % 2) {
    cout << 0 << "\n";
    return 0;
  }
  if ((s - x) / 2 & x) {
    cout << 0 << "\n";
    return 0;
  }
  long long int one = countSetBits(x);
  long long int way = pow(2, one);
  if (s == x) way -= 2;
  cout << way << "\n";
  return 0;
}

### 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;
long long int countSetBits(long long int n) {
  if (n == 0)
    return 0;
  else
    return (n & 1) + countSetBits(n >> 1);
}
int main() {
  long long int s, x;
  cin >> s >> x;
  if (x > s || (s - x) % 2) {
    cout << 0 << "\n";
    return 0;
  }
  if ((s - x) / 2 & x) {
    cout << 0 << "\n";
    return 0;
  }
  long long int one = countSetBits(x);
  long long int way = pow(2, one);
  if (s == x) way -= 2;
  cout << way << "\n";
  return 0;
}

```