#include <bits/stdc++.h>
using namespace std;
int main() {
  uint64_t sum, x;
  cin >> sum >> x;
  if (((sum - x) % 2) != 0) {
    cout << 0 << endl;
    return 0;
  }
  uint64_t a = (sum - x) / 2;
  bool correction = (a == 0);
  int i = 0;
  uint64_t temp = x;
  while (a > 0) {
    if ((a & 1) && (temp & 1)) {
      cout << 0 << endl;
      return 0;
    }
    a >>= 1;
    temp >>= 1;
  }
  uint64_t bits = 0;
  temp = 0;
  while (x != 0) {
    if (x & 1) {
      bits++;
    }
    x >>= 1;
  }
  uint64_t ans = (1LL << bits);
  if (correction) {
    cout << ans - 2 << endl;
  } else {
    cout << ans << endl;
  }
  return 0;
}

### Prompt

Create a solution in CPP for 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() {
  uint64_t sum, x;
  cin >> sum >> x;
  if (((sum - x) % 2) != 0) {
    cout << 0 << endl;
    return 0;
  }
  uint64_t a = (sum - x) / 2;
  bool correction = (a == 0);
  int i = 0;
  uint64_t temp = x;
  while (a > 0) {
    if ((a & 1) && (temp & 1)) {
      cout << 0 << endl;
      return 0;
    }
    a >>= 1;
    temp >>= 1;
  }
  uint64_t bits = 0;
  temp = 0;
  while (x != 0) {
    if (x & 1) {
      bits++;
    }
    x >>= 1;
  }
  uint64_t ans = (1LL << bits);
  if (correction) {
    cout << ans - 2 << endl;
  } else {
    cout << ans << endl;
  }
  return 0;
}

```