#include <bits/stdc++.h>
using namespace std;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  long long s, x;
  cin >> s >> x;
  if ((s - x) % 2 != 0) {
    cout << 0 << endl;
    return 0;
  }
  long long AandB = (s - x) / 2;
  if (AandB < 0 || (AandB & x) != 0) {
    cout << 0 << endl;
    return 0;
  }
  long long ones = 0;
  while (x > 0) {
    x = x & (x - 1);
    ones++;
  }
  long long ans = 1;
  ans <<= ones;
  if (AandB == 0) ans -= 2;
  cout << ans << endl;
  return 0;
}

### Prompt

Generate a Cpp solution to 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(nullptr);
  long long s, x;
  cin >> s >> x;
  if ((s - x) % 2 != 0) {
    cout << 0 << endl;
    return 0;
  }
  long long AandB = (s - x) / 2;
  if (AandB < 0 || (AandB & x) != 0) {
    cout << 0 << endl;
    return 0;
  }
  long long ones = 0;
  while (x > 0) {
    x = x & (x - 1);
    ones++;
  }
  long long ans = 1;
  ans <<= ones;
  if (AandB == 0) ans -= 2;
  cout << ans << endl;
  return 0;
}

```