#include <bits/stdc++.h>
using namespace std;
int main() {
  long long s, xo;
  cin >> s >> xo;
  long long an = (s - xo) / 2;
  if (an < 0 || an * 2 != s - xo || (an & xo) != 0) {
    cout << "0\n";
    return 0;
  }
  long long x = xo;
  long long ans = 0;
  int kol = 0;
  while (x) {
    kol += (x & 1);
    x /= 2;
  }
  ans = 1LL << kol;
  if (an == 0) {
    ans -= 2;
  }
  cout << ans << 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;
int main() {
  long long s, xo;
  cin >> s >> xo;
  long long an = (s - xo) / 2;
  if (an < 0 || an * 2 != s - xo || (an & xo) != 0) {
    cout << "0\n";
    return 0;
  }
  long long x = xo;
  long long ans = 0;
  int kol = 0;
  while (x) {
    kol += (x & 1);
    x /= 2;
  }
  ans = 1LL << kol;
  if (an == 0) {
    ans -= 2;
  }
  cout << ans << endl;
  return 0;
}

```