#include <bits/stdc++.h>
using namespace std;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long x, y;
  cin >> x >> y;
  if (x < y || (x - y) % 2)
    cout << 0 << '\n';
  else {
    bool ok = 1;
    long long r = (x - y) / 2;
    long long ans = 1;
    for (int i = (0); i < (40); i++) {
      long long pot2 = 1LL << i;
      if ((y & pot2)) {
        if (r & pot2)
          ok = 0;
        else
          ans *= 2;
      }
    }
    if (ok)
      cout << ans - 2 * (x == y) << '\n';
    else
      cout << 0 << '\n';
  }
  return 0;
}

### Prompt

Your challenge is to write 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() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long x, y;
  cin >> x >> y;
  if (x < y || (x - y) % 2)
    cout << 0 << '\n';
  else {
    bool ok = 1;
    long long r = (x - y) / 2;
    long long ans = 1;
    for (int i = (0); i < (40); i++) {
      long long pot2 = 1LL << i;
      if ((y & pot2)) {
        if (r & pot2)
          ok = 0;
        else
          ans *= 2;
      }
    }
    if (ok)
      cout << ans - 2 * (x == y) << '\n';
    else
      cout << 0 << '\n';
  }
  return 0;
}

```