#include <bits/stdc++.h>
const long long N = 250;
long long ss, aa;
long long s[N];
long long a[N];
long long f[N][2];
void init() { std::cin >> ss >> aa; }
void work() {
  long long answer = 0;
  if (ss == aa) {
    answer -= 2;
  }
  for (long long i = 1; i <= 100; i++) {
    s[i] = ss & 1;
    ss >>= 1;
    a[i] = aa & 1;
    aa >>= 1;
  }
  f[0][0] = 1;
  for (long long i = 1; i <= 200; i++) {
    long long t[2];
    for (t[0] = 0; t[0] <= 1; t[0]++) {
      t[1] = a[i] ^ t[0];
      for (long long j = 0; j <= 1; j++) {
        if (((t[0] + t[1] + j) & 1) == s[i]) {
          f[i][(t[0] + t[1] + j) >> 1] += f[i - 1][j];
        }
      }
    }
  }
  answer += f[200][0];
  std::cout << answer << std::endl;
}
int main() {
  init();
  work();
  return 0;
}

### Prompt

Develop a solution in Cpp to the problem described below:

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>
const long long N = 250;
long long ss, aa;
long long s[N];
long long a[N];
long long f[N][2];
void init() { std::cin >> ss >> aa; }
void work() {
  long long answer = 0;
  if (ss == aa) {
    answer -= 2;
  }
  for (long long i = 1; i <= 100; i++) {
    s[i] = ss & 1;
    ss >>= 1;
    a[i] = aa & 1;
    aa >>= 1;
  }
  f[0][0] = 1;
  for (long long i = 1; i <= 200; i++) {
    long long t[2];
    for (t[0] = 0; t[0] <= 1; t[0]++) {
      t[1] = a[i] ^ t[0];
      for (long long j = 0; j <= 1; j++) {
        if (((t[0] + t[1] + j) & 1) == s[i]) {
          f[i][(t[0] + t[1] + j) >> 1] += f[i - 1][j];
        }
      }
    }
  }
  answer += f[200][0];
  std::cout << answer << std::endl;
}
int main() {
  init();
  work();
  return 0;
}

```