#include <bits/stdc++.h>
using namespace std;
int main() {
  long long int s, x, y, z, c;
  int i;
  long long int ans = 1;
  bool b = false;
  cin >> s >> x;
  if ((s - x) % 2 != 0) {
    cout << 0;
    return 0;
  }
  if (s == x) {
    b = true;
  }
  z = ((s - x) >> 1);
  if ((x & z) != 0) {
    cout << 0;
    return 0;
  }
  while (x != 0) {
    ans *= (x % 2 + 1);
    x = x >> 1;
  }
  if (b) ans -= 2;
  cout << ans;
  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() {
  long long int s, x, y, z, c;
  int i;
  long long int ans = 1;
  bool b = false;
  cin >> s >> x;
  if ((s - x) % 2 != 0) {
    cout << 0;
    return 0;
  }
  if (s == x) {
    b = true;
  }
  z = ((s - x) >> 1);
  if ((x & z) != 0) {
    cout << 0;
    return 0;
  }
  while (x != 0) {
    ans *= (x % 2 + 1);
    x = x >> 1;
  }
  if (b) ans -= 2;
  cout << ans;
  return 0;
}

```