#include <bits/stdc++.h>
using namespace std;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(NULL);
  long long i, j, n, m, s, x;
  cin >> s >> x;
  long long exc = (s - x) / 2;
  if (exc < 0 || (s - x) % 2 != 0) {
    cout << 0;
    return 0;
  }
  long long cnt = 1;
  long long tot = 0;
  if (exc == 0) tot = -2;
  while (x > 0 || exc > 0) {
    long long e1 = x % 2;
    long long e2 = exc % 2;
    if (e1 == 1) {
      if (e2 == 1)
        cnt = 0;
      else
        cnt *= 2;
    }
    x /= 2;
    exc /= 2;
  }
  cnt += tot;
  cout << cnt;
  return 0;
}

### 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;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(NULL);
  long long i, j, n, m, s, x;
  cin >> s >> x;
  long long exc = (s - x) / 2;
  if (exc < 0 || (s - x) % 2 != 0) {
    cout << 0;
    return 0;
  }
  long long cnt = 1;
  long long tot = 0;
  if (exc == 0) tot = -2;
  while (x > 0 || exc > 0) {
    long long e1 = x % 2;
    long long e2 = exc % 2;
    if (e1 == 1) {
      if (e2 == 1)
        cnt = 0;
      else
        cnt *= 2;
    }
    x /= 2;
    exc /= 2;
  }
  cnt += tot;
  cout << cnt;
  return 0;
}

```