#include <bits/stdc++.h>
using namespace std;
long long solve(long long s, long long x) {
  if (s < 0)
    return 0;
  else if (s == 0) {
    return x ? 0 : 1;
  }
  if (s & 1) {
    if (x & 1) {
      return 2 * solve(s >> 1, x >> 1);
    } else {
      return 0;
    }
  } else {
    if (x & 1) {
      return 0;
    } else {
      return solve((s >> 1) - 1, x >> 1) + solve(s >> 1, x >> 1);
    }
  }
}
int main() {
  long long s, x;
  while (~scanf("%I64d%I64d", &s, &x)) {
    printf("%I64d\n", s == x ? solve(s, x) - 2 : solve(s, x));
  }
  return 0;
}

### Prompt

Please create a solution in Cpp 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;
long long solve(long long s, long long x) {
  if (s < 0)
    return 0;
  else if (s == 0) {
    return x ? 0 : 1;
  }
  if (s & 1) {
    if (x & 1) {
      return 2 * solve(s >> 1, x >> 1);
    } else {
      return 0;
    }
  } else {
    if (x & 1) {
      return 0;
    } else {
      return solve((s >> 1) - 1, x >> 1) + solve(s >> 1, x >> 1);
    }
  }
}
int main() {
  long long s, x;
  while (~scanf("%I64d%I64d", &s, &x)) {
    printf("%I64d\n", s == x ? solve(s, x) - 2 : solve(s, x));
  }
  return 0;
}

```