#include <bits/stdc++.h>
using namespace std;
long long s, x, a, a1, s1, k, ans;
long long st(long long a, long long b) {
  long long ans = 1;
  for (int i = 1; i <= b; i++) ans *= a;
  return ans;
}
int main() {
  cin >> s >> x;
  if ((s - x) % 2 != 0) {
    cout << 0;
    return 0;
  }
  a = (s - x) / 2;
  a1 = a;
  s1 = x;
  while (a1 != 0 || s1 != 0) {
    if ((s1 & 1) == 1 && (a1 & 1) == 1) {
      cout << 0;
      return 0;
    }
    a1 >>= 1;
    s1 >>= 1;
  }
  s1 = x;
  while (s1 != 0) {
    if ((s1 & 1) == 1) k++;
    s1 >>= 1;
  }
  ans = st(2, k);
  if (a == 0) ans -= 2;
  cout << ans;
  return 0;
}

### Prompt

In cpp, your task is to solve 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 s, x, a, a1, s1, k, ans;
long long st(long long a, long long b) {
  long long ans = 1;
  for (int i = 1; i <= b; i++) ans *= a;
  return ans;
}
int main() {
  cin >> s >> x;
  if ((s - x) % 2 != 0) {
    cout << 0;
    return 0;
  }
  a = (s - x) / 2;
  a1 = a;
  s1 = x;
  while (a1 != 0 || s1 != 0) {
    if ((s1 & 1) == 1 && (a1 & 1) == 1) {
      cout << 0;
      return 0;
    }
    a1 >>= 1;
    s1 >>= 1;
  }
  s1 = x;
  while (s1 != 0) {
    if ((s1 & 1) == 1) k++;
    s1 >>= 1;
  }
  ans = st(2, k);
  if (a == 0) ans -= 2;
  cout << ans;
  return 0;
}

```