#include <bits/stdc++.h>
using namespace std;
template <typename TH>
void debug_vars(const char* data, TH head) {
  cerr << data << "=" << head << "\n";
}
template <typename TH, typename... TA>
void debug_vars(const char* data, TH head, TA... tail) {
  while (*data != ',') cerr << *data++;
  cerr << "=" << head << ",";
  debug_vars(data + 1, tail...);
}
long long s, x;
void input() { cin >> s >> x; }
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  input();
  long long res = 1, mask = 0;
  for (long long a = 1; a < (1LL << 50); a *= 2) {
    if (x & a) {
      res <<= 1;
    } else {
      mask |= (2 * a);
    }
  }
  if (s < x) {
    cout << "0\n";
    return 0;
  }
  if (s == x) {
    res -= 2;
  }
  if (((s - x) | mask) == mask) {
    cout << res << "\n";
  } else {
    cout << 0 << "\n";
  }
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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;
template <typename TH>
void debug_vars(const char* data, TH head) {
  cerr << data << "=" << head << "\n";
}
template <typename TH, typename... TA>
void debug_vars(const char* data, TH head, TA... tail) {
  while (*data != ',') cerr << *data++;
  cerr << "=" << head << ",";
  debug_vars(data + 1, tail...);
}
long long s, x;
void input() { cin >> s >> x; }
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  input();
  long long res = 1, mask = 0;
  for (long long a = 1; a < (1LL << 50); a *= 2) {
    if (x & a) {
      res <<= 1;
    } else {
      mask |= (2 * a);
    }
  }
  if (s < x) {
    cout << "0\n";
    return 0;
  }
  if (s == x) {
    res -= 2;
  }
  if (((s - x) | mask) == mask) {
    cout << res << "\n";
  } else {
    cout << 0 << "\n";
  }
}

```