#include <bits/stdc++.h>
using namespace std;
long long counter(long long a, long long b) {
  if (b < 0) {
    return 0;
  }
  if ((a == 0) && (b == 0)) {
    return 1;
  }
  if (((a % 2) == 1) && ((b % 2) == 0)) {
    return 0;
  }
  if (((a % 2) == 1) && ((b % 2) == 1)) {
    return 2 * counter(a / 2, b / 2);
  }
  if (((a % 2) == 0) && ((b % 2) == 1)) {
    return 0;
  }
  if (((a % 2) == 0) && ((b % 2) == 0)) {
    return counter(a / 2, b / 2) + counter(a / 2, (b - 1) / 2);
  }
}
int main() {
  long long a, b;
  cin >> b >> a;
  long long cnt = counter(a, b);
  if (a == b) {
    cnt -= 2;
  }
  cout << cnt;
  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 counter(long long a, long long b) {
  if (b < 0) {
    return 0;
  }
  if ((a == 0) && (b == 0)) {
    return 1;
  }
  if (((a % 2) == 1) && ((b % 2) == 0)) {
    return 0;
  }
  if (((a % 2) == 1) && ((b % 2) == 1)) {
    return 2 * counter(a / 2, b / 2);
  }
  if (((a % 2) == 0) && ((b % 2) == 1)) {
    return 0;
  }
  if (((a % 2) == 0) && ((b % 2) == 0)) {
    return counter(a / 2, b / 2) + counter(a / 2, (b - 1) / 2);
  }
}
int main() {
  long long a, b;
  cin >> b >> a;
  long long cnt = counter(a, b);
  if (a == b) {
    cnt -= 2;
  }
  cout << cnt;
  return 0;
}

```