#include <bits/stdc++.h>
using namespace std;
bitset<50> xoo, soo;
long long zero, jeden, x, s, ans;
bool jest = 0, byl;
int main() {
  ios_base::sync_with_stdio(0);
  cin >> s >> x;
  xoo = x;
  soo = s;
  zero = 1;
  jeden = 0;
  for (int a = 0; a < 43; a++) {
    if (xoo[a] == 0 && soo[a] == 0) {
      jest = 1;
      jeden = zero;
    }
    if (xoo[a] == 1 && soo[a] == 1) {
      zero *= 2;
      jeden = 0;
    }
    if (xoo[a] == 1 && soo[a] == 0) {
      jest = 1;
      jeden *= 2;
      zero = 0;
    }
    if (xoo[a] == 0 && soo[a] == 1) {
      jest = 1;
      zero = jeden;
    }
    if (soo[a] || xoo[a]) ans = zero, byl = jest;
  }
  if (x == s) ans -= 2;
  cout << ans;
  return 0;
}

### 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;
bitset<50> xoo, soo;
long long zero, jeden, x, s, ans;
bool jest = 0, byl;
int main() {
  ios_base::sync_with_stdio(0);
  cin >> s >> x;
  xoo = x;
  soo = s;
  zero = 1;
  jeden = 0;
  for (int a = 0; a < 43; a++) {
    if (xoo[a] == 0 && soo[a] == 0) {
      jest = 1;
      jeden = zero;
    }
    if (xoo[a] == 1 && soo[a] == 1) {
      zero *= 2;
      jeden = 0;
    }
    if (xoo[a] == 1 && soo[a] == 0) {
      jest = 1;
      jeden *= 2;
      zero = 0;
    }
    if (xoo[a] == 0 && soo[a] == 1) {
      jest = 1;
      zero = jeden;
    }
    if (soo[a] || xoo[a]) ans = zero, byl = jest;
  }
  if (x == s) ans -= 2;
  cout << ans;
  return 0;
}

```