#include <bits/stdc++.h>
using namespace std;
signed main() {
  long long n, m;
  cin >> n >> m;
  if ((n - m) % 2 == 1 || m > n) {
    cout << 0 << endl;
    return 0;
  }
  long long ad = (n - m) / 2;
  long long oo = n - (n - m) / 2;
  long long res = oo - ad;
  long long resp = 1;
  if (ad == oo) {
    cout << 1 << endl;
    return 0;
  }
  while (res > 0) {
    resp *= 2;
    res -= (res & -res);
  }
  if ((ad & m) != 0) {
    cout << 0 << endl;
    return 0;
  }
  if (ad == 0) resp -= 2;
  if (resp == 1)
    cout << 0 << endl;
  else
    cout << resp << endl;
}

### 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;
signed main() {
  long long n, m;
  cin >> n >> m;
  if ((n - m) % 2 == 1 || m > n) {
    cout << 0 << endl;
    return 0;
  }
  long long ad = (n - m) / 2;
  long long oo = n - (n - m) / 2;
  long long res = oo - ad;
  long long resp = 1;
  if (ad == oo) {
    cout << 1 << endl;
    return 0;
  }
  while (res > 0) {
    resp *= 2;
    res -= (res & -res);
  }
  if ((ad & m) != 0) {
    cout << 0 << endl;
    return 0;
  }
  if (ad == 0) resp -= 2;
  if (resp == 1)
    cout << 0 << endl;
  else
    cout << resp << endl;
}

```