#include <bits/stdc++.h>
const int MOD = 1000000007;
using namespace std;
unsigned long long a, b, second, x, ans;
unsigned long long get(bool bit, unsigned long long second,
                       unsigned long long x) {
  if (second == 0)
    if (bit)
      return 0;
    else
      return 1;
  if (second & 1) {
    if (x & 1)
      if (bit)
        return 0;
      else
        return 2 * get(0, second / 2, x / 2);
    else if (bit)
      return get(1, second / 2, x / 2) + get(0, second / 2, x / 2);
    else
      return 0;
  } else {
    if (x & 1)
      if (bit)
        return 2 * get(1, second / 2, x / 2);
      else
        return 0;
    else if (bit)
      return 0;
    else
      return get(0, second / 2, x / 2) + get(1, second / 2, x / 2);
  }
}
int main() {
  cin >> second >> x;
  if (x > second) return cout << 0, 0;
  ans = get(0, second, x);
  if (second == x) ans -= 2;
  cout << ans;
  return 0;
}

### Prompt

Please formulate 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>
const int MOD = 1000000007;
using namespace std;
unsigned long long a, b, second, x, ans;
unsigned long long get(bool bit, unsigned long long second,
                       unsigned long long x) {
  if (second == 0)
    if (bit)
      return 0;
    else
      return 1;
  if (second & 1) {
    if (x & 1)
      if (bit)
        return 0;
      else
        return 2 * get(0, second / 2, x / 2);
    else if (bit)
      return get(1, second / 2, x / 2) + get(0, second / 2, x / 2);
    else
      return 0;
  } else {
    if (x & 1)
      if (bit)
        return 2 * get(1, second / 2, x / 2);
      else
        return 0;
    else if (bit)
      return 0;
    else
      return get(0, second / 2, x / 2) + get(1, second / 2, x / 2);
  }
}
int main() {
  cin >> second >> x;
  if (x > second) return cout << 0, 0;
  ans = get(0, second, x);
  if (second == x) ans -= 2;
  cout << ans;
  return 0;
}

```