#include <bits/stdc++.h>
using namespace std;
int main() {
  std::ios::sync_with_stdio(false);
  long long s, x;
  scanf("%lld%lld", &s, &x);
  if ((s - x) & 1ll) {
    puts("0");
    return 0;
  }
  long long a = (s - x) / 2;
  long long ans = 0;
  int b = 0;
  long long i = 40;
  for (; i >= 0; i--) {
    if (x & (1ll << i)) {
      if (a & (1ll << i)) {
        puts("0");
        return 0;
      } else
        ans++;
    } else if (a & (1ll << i))
      b++;
  }
  printf("%lld\n", (1ll << ans) - (b == 0 && ans > 0 ? 2 : 0));
  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;
int main() {
  std::ios::sync_with_stdio(false);
  long long s, x;
  scanf("%lld%lld", &s, &x);
  if ((s - x) & 1ll) {
    puts("0");
    return 0;
  }
  long long a = (s - x) / 2;
  long long ans = 0;
  int b = 0;
  long long i = 40;
  for (; i >= 0; i--) {
    if (x & (1ll << i)) {
      if (a & (1ll << i)) {
        puts("0");
        return 0;
      } else
        ans++;
    } else if (a & (1ll << i))
      b++;
  }
  printf("%lld\n", (1ll << ans) - (b == 0 && ans > 0 ? 2 : 0));
  return 0;
}

```