#include <bits/stdc++.h>
using namespace std;
long long s, x, ans = 1;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
  cin >> s >> x;
  if ((s - x) % 2) return cout << 0 << endl, 0;
  ;
  s = (s - x) / 2;
  int cnt = 0;
  for (int i = 0; i < 41; i++) {
    int one = (s >> i) & 1, two = (x >> i) & 1;
    if (one == 1 && two == 1) return cout << 0 << endl, 0;
    ;
    if (one == 0 && two == 1)
      ans *= 2;
    else if (one || two)
      cnt++;
  }
  cout << ans - (!cnt) * 2 << endl;
  return 0;
}

### Prompt

In CPP, your task is to solve 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;
long long s, x, ans = 1;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
  cin >> s >> x;
  if ((s - x) % 2) return cout << 0 << endl, 0;
  ;
  s = (s - x) / 2;
  int cnt = 0;
  for (int i = 0; i < 41; i++) {
    int one = (s >> i) & 1, two = (x >> i) & 1;
    if (one == 1 && two == 1) return cout << 0 << endl, 0;
    ;
    if (one == 0 && two == 1)
      ans *= 2;
    else if (one || two)
      cnt++;
  }
  cout << ans - (!cnt) * 2 << endl;
  return 0;
}

```