#include <bits/stdc++.h>
using namespace std;
map<pair<long long int, long long int>, long long int> m;
long long int fn(long long int s, long long int x) {
  if (s == 0 && x == 0) return 1;
  int a = s % 2, b = x % 2;
  if (a ^ b == 1) return 0;
  if (m[pair<long long int, long long int>(s, x)])
    return m[pair<long long int, long long int>(s, x)];
  s /= 2;
  x /= 2;
  if (a)
    return m[pair<long long int, long long int>(s, x)] = 2 * fn(s, x);
  else
    return m[pair<long long int, long long int>(s, x)] =
               fn(s, x) + fn(s - 1, x);
}
int main() {
  long long int s, x, ans;
  cin >> s >> x;
  ans = fn(s, x);
  if (s == x) 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;
map<pair<long long int, long long int>, long long int> m;
long long int fn(long long int s, long long int x) {
  if (s == 0 && x == 0) return 1;
  int a = s % 2, b = x % 2;
  if (a ^ b == 1) return 0;
  if (m[pair<long long int, long long int>(s, x)])
    return m[pair<long long int, long long int>(s, x)];
  s /= 2;
  x /= 2;
  if (a)
    return m[pair<long long int, long long int>(s, x)] = 2 * fn(s, x);
  else
    return m[pair<long long int, long long int>(s, x)] =
               fn(s, x) + fn(s - 1, x);
}
int main() {
  long long int s, x, ans;
  cin >> s >> x;
  ans = fn(s, x);
  if (s == x) ans -= 2;
  cout << ans;
  return 0;
}

```