#include <bits/stdc++.h>
using namespace std;
int main() {
  ios_base::sync_with_stdio(false);
  long long s, x;
  long long res;
  int i, j, k;
  bool c, equal;
  bool a[44], b[44];
  for (i = 0; i < 44; i++) {
    a[i] = false;
    b[i] = false;
  }
  cin >> s >> x;
  equal = (s == x);
  i = 0;
  while (s > 0) {
    a[i++] = ((s % 2) == 1);
    s >>= 1;
  }
  j = 0;
  while (x > 0) {
    b[j++] = ((x % 2) == 1);
    x >>= 1;
  }
  if (j > i) {
    cout << "0" << endl;
    return 0;
  }
  res = 1;
  c = false;
  while (i >= 0) {
    if (b[i]) res *= 2;
    if (c && a[i] && b[i]) {
      cout << "0" << endl;
      return 0;
    }
    if (!c && !a[i] && b[i]) {
      cout << "0" << endl;
      return 0;
    }
    c = (a[i] != b[i]);
    i--;
  }
  if (equal) res -= 2;
  cout.precision(44);
  if (c)
    cout << "0" << endl;
  else
    cout << fixed << res << endl;
  return 0;
}

### Prompt

Please formulate 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;
int main() {
  ios_base::sync_with_stdio(false);
  long long s, x;
  long long res;
  int i, j, k;
  bool c, equal;
  bool a[44], b[44];
  for (i = 0; i < 44; i++) {
    a[i] = false;
    b[i] = false;
  }
  cin >> s >> x;
  equal = (s == x);
  i = 0;
  while (s > 0) {
    a[i++] = ((s % 2) == 1);
    s >>= 1;
  }
  j = 0;
  while (x > 0) {
    b[j++] = ((x % 2) == 1);
    x >>= 1;
  }
  if (j > i) {
    cout << "0" << endl;
    return 0;
  }
  res = 1;
  c = false;
  while (i >= 0) {
    if (b[i]) res *= 2;
    if (c && a[i] && b[i]) {
      cout << "0" << endl;
      return 0;
    }
    if (!c && !a[i] && b[i]) {
      cout << "0" << endl;
      return 0;
    }
    c = (a[i] != b[i]);
    i--;
  }
  if (equal) res -= 2;
  cout.precision(44);
  if (c)
    cout << "0" << endl;
  else
    cout << fixed << res << endl;
  return 0;
}

```