#include <bits/stdc++.h>
using namespace std;
int main() {
  ios_base::sync_with_stdio(false);
  long long n, m;
  cin >> n >> m;
  if (m > n || n % 2 != m % 2) {
    cout << 0;
    return 0;
  }
  n -= m;
  n /= 2;
  if (n & m) {
    cout << 0;
    return 0;
  }
  long long f = 1;
  for (int i = 0; i < 60; i++) {
    if (m & (1ll << i)) f *= 2;
  }
  if (n + m == m)
    cout << f - 2;
  else
    cout << f;
}

### 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() {
  ios_base::sync_with_stdio(false);
  long long n, m;
  cin >> n >> m;
  if (m > n || n % 2 != m % 2) {
    cout << 0;
    return 0;
  }
  n -= m;
  n /= 2;
  if (n & m) {
    cout << 0;
    return 0;
  }
  long long f = 1;
  for (int i = 0; i < 60; i++) {
    if (m & (1ll << i)) f *= 2;
  }
  if (n + m == m)
    cout << f - 2;
  else
    cout << f;
}

```