#include <bits/stdc++.h>
using namespace std;
int main() {
  long long xsum, xxor;
  cin >> xsum >> xxor;
  long long a = (xsum - xxor) >> 1LL;
  long long b = xsum - a;
  if (xsum >= xxor && (a ^ b) == xxor) {
    long long ans = 1LL << __builtin_popcountll(xxor);
    if (xsum == xxor) {
      ans -= 2;
    }
    cout << ans << endl;
  } else {
    cout << 0 << endl;
  }
  return 0;
}

### Prompt

Develop a solution in cpp to the problem described below:

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 xsum, xxor;
  cin >> xsum >> xxor;
  long long a = (xsum - xxor) >> 1LL;
  long long b = xsum - a;
  if (xsum >= xxor && (a ^ b) == xxor) {
    long long ans = 1LL << __builtin_popcountll(xxor);
    if (xsum == xxor) {
      ans -= 2;
    }
    cout << ans << endl;
  } else {
    cout << 0 << endl;
  }
  return 0;
}

```