#include <bits/stdc++.h>
using namespace std;
const long long MOD = 1e9 + 7LL;
const int MAX = 2009;
int main() {
  long long sum, xor_, and_, ans = 1;
  cin >> sum >> xor_;
  long long doubleAnd = sum - xor_;
  if (doubleAnd % 2 == 1 || sum < xor_) {
    cout << 0 << endl;
    return 0;
  }
  and_ = doubleAnd / 2;
  long long xor_cpy = xor_;
  while (and_ > 0 || xor_ > 0) {
    if (xor_ & 1 && and_ & 1) {
      cout << 0 << endl;
      return 0;
    } else if (xor_ & 1) {
      ans *= 2;
    }
    and_ >>= 1;
    xor_ >>= 1;
  }
  if (xor_cpy == sum) {
    ans -= 2;
  }
  cout << ans << endl;
  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;
const long long MOD = 1e9 + 7LL;
const int MAX = 2009;
int main() {
  long long sum, xor_, and_, ans = 1;
  cin >> sum >> xor_;
  long long doubleAnd = sum - xor_;
  if (doubleAnd % 2 == 1 || sum < xor_) {
    cout << 0 << endl;
    return 0;
  }
  and_ = doubleAnd / 2;
  long long xor_cpy = xor_;
  while (and_ > 0 || xor_ > 0) {
    if (xor_ & 1 && and_ & 1) {
      cout << 0 << endl;
      return 0;
    } else if (xor_ & 1) {
      ans *= 2;
    }
    and_ >>= 1;
    xor_ >>= 1;
  }
  if (xor_cpy == sum) {
    ans -= 2;
  }
  cout << ans << endl;
  return 0;
}

```