#include <bits/stdc++.h>
using namespace std;
const int MASK = (1 << 20) - 1;
int main() {
  long long s, x;
  cin >> s >> x;
  int count1 = 0, count2 = 0;
  int count3 = 0, count4 = 0;
  for (int a = 0; a < 1 << 20; ++a) {
    int b = (int)((x ^ a) & MASK);
    if (((a + b) & MASK) == (int)(s & MASK)) {
      if (a + b <= MASK) {
        ++count1;
      } else {
        ++count2;
      }
    }
    b = (int)((x >> 20) ^ a);
    if (a + b == (int)(s >> 20)) {
      ++count3;
    } else if (a + b + 1 == (int)(s >> 20)) {
      ++count4;
    }
  }
  long long result = count1 * (long long)count3 + count2 * (long long)count4;
  if (s == x) result -= 2;
  cout << result << endl;
  return 0;
}

### Prompt

Please create a solution in cpp 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 int MASK = (1 << 20) - 1;
int main() {
  long long s, x;
  cin >> s >> x;
  int count1 = 0, count2 = 0;
  int count3 = 0, count4 = 0;
  for (int a = 0; a < 1 << 20; ++a) {
    int b = (int)((x ^ a) & MASK);
    if (((a + b) & MASK) == (int)(s & MASK)) {
      if (a + b <= MASK) {
        ++count1;
      } else {
        ++count2;
      }
    }
    b = (int)((x >> 20) ^ a);
    if (a + b == (int)(s >> 20)) {
      ++count3;
    } else if (a + b + 1 == (int)(s >> 20)) {
      ++count4;
    }
  }
  long long result = count1 * (long long)count3 + count2 * (long long)count4;
  if (s == x) result -= 2;
  cout << result << endl;
  return 0;
}

```