#include <bits/stdc++.h>
using namespace std;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  long long s, x;
  cin >> s >> x;
  if (s < x || (s % 2) != (x % 2)) {
    cout << "0\n";
    return 0;
  }
  long long a = (s - x) / 2, res = 1;
  if (a & x) {
    cout << "0\n";
    return 0;
  }
  for (long long b = 0; b < 50; b++) {
    if (((1LL << b) & a) == 0 && ((1LL << b) & x) != 0) res *= 2;
  }
  if (s == x) res -= 2;
  cout << res << '\n';
}

### Prompt

Please formulate a cpp solution 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);
  cin.tie(0);
  long long s, x;
  cin >> s >> x;
  if (s < x || (s % 2) != (x % 2)) {
    cout << "0\n";
    return 0;
  }
  long long a = (s - x) / 2, res = 1;
  if (a & x) {
    cout << "0\n";
    return 0;
  }
  for (long long b = 0; b < 50; b++) {
    if (((1LL << b) & a) == 0 && ((1LL << b) & x) != 0) res *= 2;
  }
  if (s == x) res -= 2;
  cout << res << '\n';
}

```