#include <bits/stdc++.h>
using namespace std;
typedef int64_t ll;
typedef uint64_t ull;
int main(int argc, char **argv) {
  ios_base::sync_with_stdio(false);
  cin.tie(0), cout.tie(0), cout.precision(15);
  ll s, x;
  cin >> s >> x;
  ll carry = s - x;
  if (carry & 1) {
    cout << 0 << endl;
    return 0;
  }
  carry >>= 1;
  int cnt = 0;
  for (int i = 0; i < 63; i++) {
    ull bmask = (1ULL << i);
    if (x & bmask) {
      if (carry & bmask) {
        cout << 0 << endl;
        return 0;
      }
      cnt++;
    }
  }
  int zeros = 0;
  if (s == x) {
    zeros = 2;
  }
  cout << (1ULL << cnt) - zeros << endl;
  return 0;
}

### Prompt

Create a solution in CPP for 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;
typedef int64_t ll;
typedef uint64_t ull;
int main(int argc, char **argv) {
  ios_base::sync_with_stdio(false);
  cin.tie(0), cout.tie(0), cout.precision(15);
  ll s, x;
  cin >> s >> x;
  ll carry = s - x;
  if (carry & 1) {
    cout << 0 << endl;
    return 0;
  }
  carry >>= 1;
  int cnt = 0;
  for (int i = 0; i < 63; i++) {
    ull bmask = (1ULL << i);
    if (x & bmask) {
      if (carry & bmask) {
        cout << 0 << endl;
        return 0;
      }
      cnt++;
    }
  }
  int zeros = 0;
  if (s == x) {
    zeros = 2;
  }
  cout << (1ULL << cnt) - zeros << endl;
  return 0;
}

```