#include <bits/stdc++.h>
using namespace std;
template <class T>
bool umin(T& a, T b) {
  if (a > b) {
    a = b;
    return 1;
  }
  return 0;
}
template <class T>
bool umax(T& a, T b) {
  if (a < b) {
    a = b;
    return 1;
  }
  return 0;
}
int main() {
  long long s, x;
  scanf("%lld%lld", &s, &x);
  long long a = (s - x) / 2;
  if (a < 0 or x + 2 * a != s or (x & a) != 0) return puts("0") * 0;
  long long tmp = x, res = 0;
  while (tmp > 0) {
    res += tmp & 1;
    tmp /= 2;
  }
  long long ans = (1LL << res);
  if (!a) ans -= 2;
  printf("%lld\n", ans);
  return 0;
}

### Prompt

Your task is to create a Cpp solution 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;
template <class T>
bool umin(T& a, T b) {
  if (a > b) {
    a = b;
    return 1;
  }
  return 0;
}
template <class T>
bool umax(T& a, T b) {
  if (a < b) {
    a = b;
    return 1;
  }
  return 0;
}
int main() {
  long long s, x;
  scanf("%lld%lld", &s, &x);
  long long a = (s - x) / 2;
  if (a < 0 or x + 2 * a != s or (x & a) != 0) return puts("0") * 0;
  long long tmp = x, res = 0;
  while (tmp > 0) {
    res += tmp & 1;
    tmp /= 2;
  }
  long long ans = (1LL << res);
  if (!a) ans -= 2;
  printf("%lld\n", ans);
  return 0;
}

```