#include <bits/stdc++.h>
using namespace std;
long long int po(long long int c) {
  long long int an = 1;
  for (int i = 0; i < c; i++) an *= 2;
  return an;
}
int main() {
  bool f[50];
  bool fs[50];
  for (int i = 0; i < 50; i++) {
    f[i] = false;
    fs[i] = false;
  }
  long long int s, x, t = 0;
  cin >> s >> x;
  long long int n = x, c = 0;
  while (n > 0) {
    if (n % 2 == 1) {
      f[c] = 1;
      t++;
    }
    c++;
    n /= 2;
  }
  s -= x;
  if (s < 0) {
    cout << "0";
    return 0;
  }
  if (s == 0) {
    cout << po(t) - 2;
    return 0;
  }
  if (s % 2 != 0) {
    cout << "0";
    return 0;
  }
  n = s / 2, c = 0;
  while (n > 0) {
    if (n % 2 == 1) {
      fs[c] = 1;
    }
    c++;
    n /= 2;
  }
  s /= 2;
  c = 49;
  for (int i = 0; i < 50; i++)
    if (fs[i] && f[i]) {
      cout << "0";
      return 0;
    }
  cout << po(t);
  return 0;
}

### Prompt

Your task is to create 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;
long long int po(long long int c) {
  long long int an = 1;
  for (int i = 0; i < c; i++) an *= 2;
  return an;
}
int main() {
  bool f[50];
  bool fs[50];
  for (int i = 0; i < 50; i++) {
    f[i] = false;
    fs[i] = false;
  }
  long long int s, x, t = 0;
  cin >> s >> x;
  long long int n = x, c = 0;
  while (n > 0) {
    if (n % 2 == 1) {
      f[c] = 1;
      t++;
    }
    c++;
    n /= 2;
  }
  s -= x;
  if (s < 0) {
    cout << "0";
    return 0;
  }
  if (s == 0) {
    cout << po(t) - 2;
    return 0;
  }
  if (s % 2 != 0) {
    cout << "0";
    return 0;
  }
  n = s / 2, c = 0;
  while (n > 0) {
    if (n % 2 == 1) {
      fs[c] = 1;
    }
    c++;
    n /= 2;
  }
  s /= 2;
  c = 49;
  for (int i = 0; i < 50; i++)
    if (fs[i] && f[i]) {
      cout << "0";
      return 0;
    }
  cout << po(t);
  return 0;
}

```