#include <bits/stdc++.h>
using namespace std;
int main() {
  long long s, xx, x;
  cin >> s >> xx;
  x = xx;
  long long t = (s - x) / 2;
  long long ans = 0, ans1 = 0;
  bool ok = 1, haveo = 0;
  while (t > 0 || x > 0) {
    if ((x & 1) == 1) haveo = 1;
    if ((t & 1) == 1 && (x & 1) == 1) {
      ok = 0;
      break;
    }
    if ((t & 1) == 0 && (x & 1) == 1) {
      ans++;
    }
    t >>= 1;
    x >>= 1;
  }
  if (!ok || s < xx || (s - xx) % 2 == 1)
    cout << "0" << endl;
  else {
    ans = 1ll << ans;
    if (s - xx == 0) ans -= 2;
    cout << ans << endl;
  }
  return 0;
}

### Prompt

Generate 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() {
  long long s, xx, x;
  cin >> s >> xx;
  x = xx;
  long long t = (s - x) / 2;
  long long ans = 0, ans1 = 0;
  bool ok = 1, haveo = 0;
  while (t > 0 || x > 0) {
    if ((x & 1) == 1) haveo = 1;
    if ((t & 1) == 1 && (x & 1) == 1) {
      ok = 0;
      break;
    }
    if ((t & 1) == 0 && (x & 1) == 1) {
      ans++;
    }
    t >>= 1;
    x >>= 1;
  }
  if (!ok || s < xx || (s - xx) % 2 == 1)
    cout << "0" << endl;
  else {
    ans = 1ll << ans;
    if (s - xx == 0) ans -= 2;
    cout << ans << endl;
  }
  return 0;
}

```