#include <bits/stdc++.h>
using namespace std;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  long long sum, cor, x;
  cin >> sum >> x;
  if (sum < x)
    cout << "0";
  else if (x == 0 && sum != 0)
    cout << "1";
  else {
    long long a = (sum - x) / 2, ans = 1;
    if (a < 0 || x + 2 * a != sum || (x & a) != 0) {
      cout << 0 << endl;
      return 0;
    }
    long long tmp = x;
    while (tmp > 0) {
      if (tmp & 1) ans = ans * 2;
      tmp >>= 1;
    }
    if (a == 0) ans -= 2;
    cout << ans << endl;
  }
}

### 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() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  long long sum, cor, x;
  cin >> sum >> x;
  if (sum < x)
    cout << "0";
  else if (x == 0 && sum != 0)
    cout << "1";
  else {
    long long a = (sum - x) / 2, ans = 1;
    if (a < 0 || x + 2 * a != sum || (x & a) != 0) {
      cout << 0 << endl;
      return 0;
    }
    long long tmp = x;
    while (tmp > 0) {
      if (tmp & 1) ans = ans * 2;
      tmp >>= 1;
    }
    if (a == 0) ans -= 2;
    cout << ans << endl;
  }
}

```