#include <bits/stdc++.h>
using namespace std;
signed main() {
  long long s, x;
  cin >> s >> x;
  if ((s - x) % 2 == 1) {
    cout << 0;
    return 0;
  }
  long long a = (s - x) / 2;
  long long flag = 0;
  long long y = 0;
  long long s1 = s;
  long long x1 = x;
  while (x > 0) {
    long long q = x & 1;
    long long r = a & 1;
    if (q == 1) {
      if (r == 0)
        y++;
      else
        flag = 1;
    }
    a = a >> 1;
    x = x >> 1;
  }
  if (flag == 1)
    cout << 0;
  else {
    if (s1 == x1)
      cout << (long long)pow(2, y) - 2;
    else
      cout << (long long)pow(2, y);
  }
}

### Prompt

Develop a solution in CPP to the problem described below:

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 s, x;
  cin >> s >> x;
  if ((s - x) % 2 == 1) {
    cout << 0;
    return 0;
  }
  long long a = (s - x) / 2;
  long long flag = 0;
  long long y = 0;
  long long s1 = s;
  long long x1 = x;
  while (x > 0) {
    long long q = x & 1;
    long long r = a & 1;
    if (q == 1) {
      if (r == 0)
        y++;
      else
        flag = 1;
    }
    a = a >> 1;
    x = x >> 1;
  }
  if (flag == 1)
    cout << 0;
  else {
    if (s1 == x1)
      cout << (long long)pow(2, y) - 2;
    else
      cout << (long long)pow(2, y);
  }
}

```