#include <bits/stdc++.h>
using namespace std;
const int inf = (int)1e9;
const int mod = inf + 7;
const double eps = 1e-9;
const double pi = acos(-1.0);
long long second, x;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cin >> second >> x;
  long long nx = x;
  long long res = 1;
  if ((x & 1) != (second & 1)) return cout << 0, 0;
  while (nx != 0) {
    if (nx & 1) res *= 2;
    nx /= 2;
  }
  if (second == x) {
    cout << res - 2 << endl;
    return 0;
  }
  long long first = second - x;
  if (first < 0) {
    cout << 0 << endl;
    return 0;
  }
  for (int i = 0; i < 51; ++i) {
    long long a = (1ll << i);
    long long b = (1ll << (i + 1));
    if ((x & a) > 0 && (first & b) > 0) {
      cout << 0 << endl;
      return 0;
    }
  }
  cout << res;
}

### Prompt

Your challenge is to write 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;
const int inf = (int)1e9;
const int mod = inf + 7;
const double eps = 1e-9;
const double pi = acos(-1.0);
long long second, x;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cin >> second >> x;
  long long nx = x;
  long long res = 1;
  if ((x & 1) != (second & 1)) return cout << 0, 0;
  while (nx != 0) {
    if (nx & 1) res *= 2;
    nx /= 2;
  }
  if (second == x) {
    cout << res - 2 << endl;
    return 0;
  }
  long long first = second - x;
  if (first < 0) {
    cout << 0 << endl;
    return 0;
  }
  for (int i = 0; i < 51; ++i) {
    long long a = (1ll << i);
    long long b = (1ll << (i + 1));
    if ((x & a) > 0 && (first & b) > 0) {
      cout << 0 << endl;
      return 0;
    }
  }
  cout << res;
}

```