#include <bits/stdc++.h>
using namespace std;
int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  long long n, m;
  while (cin >> n >> m) {
    long long dif = n - m;
    if (dif < 0 || (dif & 1)) {
      cout << "0" << endl;
      continue;
    }
    long long ans = 1LL;
    for (int i = (0); i <= (40); i++) {
      if (m & (1LL << i)) {
        if (dif & (1LL << (i + 1))) {
          ans = 0;
          break;
        } else {
          ans *= 2;
        }
      }
    }
    if (dif == 0) ans -= 2;
    cout << ans << endl;
  }
  return 0;
}

### Prompt

Please formulate a cpp solution to 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() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  long long n, m;
  while (cin >> n >> m) {
    long long dif = n - m;
    if (dif < 0 || (dif & 1)) {
      cout << "0" << endl;
      continue;
    }
    long long ans = 1LL;
    for (int i = (0); i <= (40); i++) {
      if (m & (1LL << i)) {
        if (dif & (1LL << (i + 1))) {
          ans = 0;
          break;
        } else {
          ans *= 2;
        }
      }
    }
    if (dif == 0) ans -= 2;
    cout << ans << endl;
  }
  return 0;
}

```