#include <bits/stdc++.h>
using namespace std;
long long S, X, ans = 1;
int main() {
  scanf("%lld", &S), scanf("%lld", &X);
  for (int i = 0; i < (int)60; i++) {
    if (X & (1ll << i)) {
      ans *= 2;
      if (((S - X) / 2) & (1ll << i)) ans = 0;
    }
  }
  if ((S - X) & 1)
    printf("0\n");
  else
    printf("%lld\n", ans - 2 * (S == X));
}

### 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 S, X, ans = 1;
int main() {
  scanf("%lld", &S), scanf("%lld", &X);
  for (int i = 0; i < (int)60; i++) {
    if (X & (1ll << i)) {
      ans *= 2;
      if (((S - X) / 2) & (1ll << i)) ans = 0;
    }
  }
  if ((S - X) & 1)
    printf("0\n");
  else
    printf("%lld\n", ans - 2 * (S == X));
}

```