#include <bits/stdc++.h>
using namespace std;
int main() {
  long long s, x, ans = 1;
  cin >> s >> x;
  long MAX = log2(max(s, x)) + 1;
  bool with_carry = false;
  for (long i = 0; i <= MAX; ++i) {
    long sbit = s >> i & 1, xbit = x >> i & 1;
    long nsbit = s >> (i + 1) & 1, nxbit = x >> (i + 1) & 1;
    if ((sbit == xbit) == with_carry) ans = 0;
    if (sbit == 1 && xbit == 1) {
      with_carry = false;
      ans *= 2;
    } else if (sbit == 1 && xbit == 0) {
      with_carry = nsbit != nxbit;
    } else if (sbit == 0 && xbit == 1) {
      with_carry = true;
      ans *= 2;
    } else if (sbit == 0 && xbit == 0) {
      with_carry = nsbit != nxbit;
    }
  }
  if (ans == 0) {
    cout << 0 << endl;
    return 0;
  }
  cout << (s == x ? ans - 2 : ans) << endl;
}

### Prompt

Please create a solution in Cpp 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() {
  long long s, x, ans = 1;
  cin >> s >> x;
  long MAX = log2(max(s, x)) + 1;
  bool with_carry = false;
  for (long i = 0; i <= MAX; ++i) {
    long sbit = s >> i & 1, xbit = x >> i & 1;
    long nsbit = s >> (i + 1) & 1, nxbit = x >> (i + 1) & 1;
    if ((sbit == xbit) == with_carry) ans = 0;
    if (sbit == 1 && xbit == 1) {
      with_carry = false;
      ans *= 2;
    } else if (sbit == 1 && xbit == 0) {
      with_carry = nsbit != nxbit;
    } else if (sbit == 0 && xbit == 1) {
      with_carry = true;
      ans *= 2;
    } else if (sbit == 0 && xbit == 0) {
      with_carry = nsbit != nxbit;
    }
  }
  if (ans == 0) {
    cout << 0 << endl;
    return 0;
  }
  cout << (s == x ? ans - 2 : ans) << endl;
}

```