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

### 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 & 1) != (x & 1)) return 0;
  if (s == 0 && x == 0) return 1;
  if (s == 1 && x == 1) return 2;
  if (s & 1) return 2 * solve(s >> 1, x >> 1);
  return solve((s >> 1) - 1, x >> 1) + solve((s >> 1), x >> 1);
}
int main() {
  long long s, x;
  scanf("%lld %lld", &s, &x);
  printf("%lld\n", solve(s, x) - 2 * (s == x));
}

```