#include <bits/stdc++.h>
using namespace std;
long long int s, x;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cin >> s >> x;
  if ((s - x) % 2 != 0 || s < x) {
    printf("0");
    return 0;
  }
  long long int pc = __builtin_popcountll(x);
  long long int ans = 1ll;
  ans <<= pc;
  long long int eq = (s - x) / 2;
  for (long long int i = 0; i < 63; i++) {
    if ((eq & (1ll << i)) && (x & (1ll << i))) {
      printf("0");
      return 0;
    }
  }
  if (eq == 0) ans -= 2;
  cout << ans << endl;
  return 0;
}

### Prompt

Please provide a Cpp coded solution 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;
long long int s, x;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cin >> s >> x;
  if ((s - x) % 2 != 0 || s < x) {
    printf("0");
    return 0;
  }
  long long int pc = __builtin_popcountll(x);
  long long int ans = 1ll;
  ans <<= pc;
  long long int eq = (s - x) / 2;
  for (long long int i = 0; i < 63; i++) {
    if ((eq & (1ll << i)) && (x & (1ll << i))) {
      printf("0");
      return 0;
    }
  }
  if (eq == 0) ans -= 2;
  cout << ans << endl;
  return 0;
}

```